Essential Astronomy & Astrophysics Through "Questions " and

Prepared for submission to JHEP

Essential Astronomy & Astrophysics

Through "Questions " and "Answers"

Salah Nasri

United Arab Emirates University,

Al-Ain, UAE

E-mail: [email protected]

Abstract:

In these notes, I discuss a number of selected of topics in Astronomy and Astro-

physics that can be of interest to both science and non-science students. For a

non-physics major student or layman, very short answers to these questions are

presented, whereas for those with special interest in physics and astrophysics more

details are provided. I also included many appendices with comprehensive discus-

sion relevant to the questions discussed in these notes.

The ﬁgures in these notes are produced by the author, unless indicated.

Contents

1 How did the ancients know that Earth was round ? 2

2 Who ﬁrst estimated the distances and sizes of the Sun and the Moon? 3

3 What is the average Earth-Sun distance in kilometers ? 9

4 What is the average Earth -Moon distance ? 12

5 How did the ancients estimate the size of the Earth? 13

6 What is the simplest clear experimental evidence for Earth rotation ? 15

7 How fast does Earth revolve around the Sun? 18

8 What is the relationship between sidereal and synodic periods ? 19

9 What is the Copernican model of the solar system ? 20

10 What is the modern view of the solar system? 24

11 What is Bode’s Law? 27

12 What is the universal law of gravitation ? 28

13 What is the minimum period of rotation a spherical astronomical object

can have ? 31

14 What is the mass of the Earth ? 32

15 What is gravity at the surface of an astronomical object ? 34

16 What is the escape velocity from the Earth? 35

17 What is the age of the Earth? 36

18 What is the critical temperature of a planet’s atmosphere above which a

particular atmospheric component will be lost into space ? 41

19 What type of electromagnetic spectrum reach the Earth’s surface ? 45

20 What are some of the astrophysical sources of radio waves ? 48

– i –

21 What is the Ozone layer ? 49

22 What is a parallax angle? a parsec? a light year ? 52

23 What is the mass of the Moon ? 55

24 What is stellar magnitude? 56

25 What is the average solar power density at the Earth’s atmosphere? 63

26 How much electromagnetic power is radiated by the Sun? 63

27 How much is the surface temperature of the Sun? 64

28 Is there water on Mars? 65

29 How do we know if a star or a galaxy is moving toward or away from us? 67

30 What is the spectral classiﬁcation of stars ? 67

31 What is the Eddington limit ? 68

32 What is the minimum temperature for hydrogen fusion inside a star ? 68

33 What is the smallest mass that a star can have? 70

34 What is the largest mass a star can have? 70

35 What is a white dwarf? 71

36 What is a black hole (BH)? 71

37 How are black hole (BH) discovered? 72

38 What is the Event Horizon Telescope (EHT)? 73

39 What is the resolving power of a telescope? 74

40 What are the major components of a radio Telescope? 74

41 What is the epoch of reionisation? 75

42 How many galaxies are there in our visible universe? 75

43 What is the closest known galaxy to the Milky Way? 75

44 What is the farthest known galaxy in the universe? 76

– ii –

45 What is a neutron star? 76

46 How old is the universe? 76

47 How big is the universe? 76

48 What is the Cosmic Microwave Background? 76

49 What is the epoch of dark ages? 77

50 What is era of big bang nucleosynthesis? 77

51 Appendices 78

51.1 Tunneling Through Coulomb Barrier 78

51.2 The Rate of Nuclear Fusion in the Core of Stars 82

51.3 Sunyaev-Zel’dovich (SZ) Eﬀect 84

51.4 Speciﬁc Intensity and Radiation Pressure 84

51.5 Radiation Transport 84

51.6 Dark Matter in Cluster of Galaxies from Hot X-ray Observations 84

52 References 85

– 1 –

1 How did the ancients know that Earth was round ?

• The idea of spherical Earth was ﬁrst proposed by Pythagoras

, however

it was Mainly based on aesthetic ground.

• It was Aristotle

who argued, based on a number of observational facts

that the shape of the Earth is spherical.

Greek philosopher who lived around 500 B.C,who made important discoveries in mathematics

and astronomy. He best known for the Pythagorean theorem.

Ancient Greek philosopher and scientist who lived between circa 384 B.C and 322 B.C, and is

considered one of the greatest thinkers in history.

More Details

The observational facts are:

• Some stars that are visible in the Northern hemisphere are not for an

observer in the Southern hemisphere. Similar argument holds the other

way around. For instance, an observer located in the Southern hemisphere can

not see the star "Polaris", but once in the Northern hemisphere it becomes

visible. Thus, the view of constellations changes when going from northern

hemisphere to the southern hemisphere, and vice versa.

• The Earth’s shadow on the moon surface during the diﬀerent phases

of the lunar eclipse always evolves from arched to circular, regardless

of the locations of the observer on the Earth

. Moreover, when the moon

was in diﬀerent positions during an eclipse, the sunlight struck the earth from

diﬀerent angles, and yet the shape of the Earth’s shadow, as it moves across

the moon, always looked part of a circle. This means that Earth has a shape

that casts a circular shadow in every possible angle. This could only be caused

by a spherical shape.

• When standing on the shore and watching a ship moving away, the bottom

of ship in sea starts to disappear ﬁrst over the horizon, and then,

gradually, the mast.

If the Earth had a shape of a disk, then the moon would have to be directly overhead to get a

round shadow, which means the sun would be directly below the earth. However, if that was the

case, the entire surface of the Earth would be dark during a lunar eclipse.

– 2 –

Figure 1: More stars are visible to an observer moving toward the north and looking

to the sky toward the North. Similarly, more stars will be visible to her/him when

moving toward the south and looking to the sky in the direction toward the south.

2 Who ﬁrst estimated the distances and sizes of the Sun and the Moon?

Quick Answer

• Around 250 B.C, Aristarchus of Samos

used observations of lunar eclipses

and simple geometry to estimate the relative distances and sizes of the

Earth, Moon, and Sun. He found that

E-M

' 10;

E-S

' 191;

' 19;

' 2.85

where,

E-M

= Earth-Moon distance, d

E-S

= Earth-Sun distance,

= Earth’s diameter, D

= Moon’s diameter, D

= Sun’s diameter

• Although his results were oﬀ by one to two orders of magnitudes from

the actual values, his method of calculation was correct.

• Equally important, his estimates showed that the universe is vast and the

Sun is much larger than the Earth.

Samos is a Greek island in the Aegean Sea.

– 3 –

More Details

• Aristarchus’s calculation of the distances and sizes of the Moon and Sun was

based on four observations. These are as follows:

Observation #1 :

"When the moon is exactly half full as seen from the Earth, the angle

between the lines of sight to the Sun and to the Moon is 87

◦

Let us denote the Earth, Moon, and Sun by the letters E, M, and S denote,

respectively, and by α the angle between the lines of sight to the Sun and

to the Moon. When the Moon is exactly half full the angle

EMS is exactly

right angle, and so by measuring the angle α it would be possible to determine

the distance to the Sun in terms of that of the Moon from the trigonometric

relation

E-M

E-S

= cos α (2.1)

How did Aristarchus determined the angle α ? He used the following simple

and cleaver idea. He calculated the angle  shown in Fig. 2 below, which is

equal to (90

◦

− α), in terms of the diﬀerence between the time the Moon takes

to go from the ﬁrst quarter to the third quarter, ∆t

(1→3)

, and the time it takes

from the third quarter to the ﬁrst quarter, ∆t

(3→1)

, given by the relation

Figure 2: Determining the angle  by measuring the time it takes the Moon to

go from the ﬁrst quarter to the third quarter and the time it takes from the third

quarter to the ﬁrst quarter. The diagram in this ﬁgure is not to scale.

∆t

(1→3)

− ∆t

(3→1)

4

2π

(2.2)

– 4 –

Here T

= 29.5 days, corresponding to the synodic orbital period of the Moon

around the Earth. Aristarchus found that (∆t

− ∆t

) = 1 day, and deduced

that

 = 90

◦



1 day

29.5 day



' 3

◦

=⇒ α ' 87

◦

(2.3)

Plugging this value into Eq. (2.1), gives



E-S

E-M



[Aristarchus]

= 19.1 (2.4)

This about a factor of 20 smaller than the actual value. The reason for this is

....

Observation #2 :

"During solar eclipse, the Sun have approximately the same angular size."

Figure 3: The Moon and the Sun have approximately the same angular size by an

observer on the Earth. The diagram in this ﬁgure is not to scale.

Mathematically this statement implies that (see Fig. 3)

E-S

E-M

(2.5)

and using the value of d

E-S

E-M

calculated ration in Eq (2.4), we get





[Aristarchus]

= 19.1 (2.6)

This value is about a factor of 20 smaller than the actual value. This discrep-

ancy is due to the underestimated value of the ratio of the Earth-Sun distance

to the Earth-Moon distance, discussed above.

– 5 –

Observation #3 :

"During lunar eclipse, the Moon travels a distance 2D

while passing the

Earth’s shadow."

Figure 4: The Moon and the Sun have approximately the same angular size by an

observer on the Earth. The diagram in this ﬁgure is not to scale.

From the similarity of the triangles ∆(abc) and ∆(cde) in Fig. 4, we have

E-M

/2 −D

shadow

E-S

/2 − D

⇐⇒

E-M

− D

shadow

E-S

− D

(2.7)

Replacing d

E-S

E-M

by the ratio D

(see Eq. (2.5)), and D

shadow

= 2D

(from observation #3), we can express the ratio D

1 + D



1 +

shadow



= 3

1 + D

(2.8)

Now we plug the value D

= 19.1 that we obtained in Eq. (2.6)) into the

above equation and ﬁnd





[Aristarchus]

= 2.85 (2.9)

Aristarchus found 2.51 < D

< 3.16, which is consistent with the above

result. The correct value of D

is 3.67, i.e 28% larger than the value

given above. The reason for the diﬀerence between the calculated value and

observation is size of the Earth’s shadow, D

shadow

, which is almost three time

the diameter of the Moon instead of two times.

Observation #4 :

– 6 –

"The Moon subtends one ﬁfteenth part of a sign of the Zodiac."

Since there are 12 zodiac signs, it implies that, according to Aristarchus, the

Moon has the angular diameter





360

◦



= 2

◦

(2.10)

With θ

being a small angle, we can approximate d

E-M

by the length of the

arc in a circle of radius D

subtended by θ

, and hence we write

= d

E-M

= d

E-M



◦

180

◦



× 3.14 rd = 0.035 d

E-M

(2.11)

With the use of this result and Eq. (2.9), we have



E-M



[Aristarchus]

E-M



0.035



2.85



= 10 (2.12)

which is about of factor 3 small than the actual value. Part of the discrepancy

is due to the fact that the value of θ

quoted by Aristarchus was largely

overestimated by a factor of four

• About century after Aristarchus, the Greek astronomer Hipparchus [190 −

120 B.C] of Nicaea used the observation of solar eclipse to determine the Earth-

Moon distance in terms of the radius of the Earth which was closer to the

actual value than the value estimated by Aristarchus. He knew that at the

city of "Hellespont", in modern day Turkey, the eclipse was total while at

Alexandria (about 500 km apart) in Egypt, the eclipse was partial, covering

about 4/5 of the Sun (i.e. partial eclipse). In Fig..., α denote the angle that

corresponds to 1/5 of the angular size of the Moon that is uncovered as viewed

from Alexandria. Hipparchus had measured the angular size of the Moon and

the Sun and found that they subtend an angle of 0.5 degree. Thus,

α =





× 0.55

◦

= 0.11

◦

(2.13)

Hipparchus also knew that the diﬀerence in latitudes between Hellespont and

Alexandria is approximately 9 degrees. Consequently (see Fig....), we have

◦

2πR

360

◦

(2.14)

2πd

E-M

360

◦

The actual angular size of the Moon (and also the Sun, from observation #2) is 0.52 degree. If

we use the correct value for the angle (i.e. θ

= 0.52), and D

= 3.67 we obtain (d

E-M

) =

29.9, which is in very good agreement with observation.

– 7 –

By solving for d

E-M

yields



E-M



[Hipparchus]



◦



= 82

Compared to the actual value, this value is overestimated by about 30%, and

the reason for this is because the Moon was assumed to be directly overhead

at the time of the eclipse, but at some signiﬁcant angle.

– 8 –

3 What is the average Earth-Sun distance in kilometers ?

Quick Answer

E−S

= 149.6 million km

More Details

• Measuring Earth-Sun distance using the transit of Venus

The transit of a planet is its passage from one end of the disk of the Sun to the

other, and it can be observed only for planets whose orbits are inside Earth’s

orbit, i.e. for the planets Mercury and Venus.

The idea of using the transits of Venus or Mercury to calculate the Earth-Sun

distance was ﬁrst proposed in 1663 by the mathematician James Gregory,

but it was the English astronomer Edmond Halley who in 1716 suggested

that of the two observing the transits of Venus will allow to determine the

distance more accuracy

. The method requires measuring the times of the be-

ginning and the end of the transit seen by two observers, A and B, from two

places on Earth separated by large latitudes

Let us denote by R



the radius of the Sun, ∆



' 15 arcmin half its angular

size, d

E−V

Venus-Earth distance, ω

= 0.067 arcsec/sec the angular velocity

of venus, ∆t

and ∆t

be the duration of the transit as measured by the ob-

servers A and B, respectively, d

the distance between their locations, θ

and

are the parallax angles measured from the locations of the two observers for

the Sun and Venus, respectively, (α

− α

) is the angular separation between

the two transits. Looking at Fig. ..., we have

E−V

=⇒ d

E−S



0.28



(3.1)

The transits of Venus are incredibly rare because the orbits of the Earth and Venus are inclined

with respect to each other, and occur in pairs about 8 years apart, and sometimes century apart.

For example: (June 1761, June 1769), (December 1874, December 1882), ect.

As we will discuss in A 16, the measurement of distance from two separated places is called

parallax method and was used by Cassini and Richter to measure the distance to Mars, from which

they inferred the Earth-Sun distance (see A 17 ).

– 9 –

Where in the second equality we made the approximation tan θ

' θ

since

E−V

much larger than the size of the Earth. We also know that d

V S

is approx-

imately 0.72 times the distance Earth-Sun

.From the triangles ACO and BCV

we see that the angle θ

can be written in terms of the angular separation as

= |α

− α

| + α



(3.2)

Since the angles α

and α

are very small

, we can approximate the angular

size of the projected paths of Venus on the Sun’s disk from the start to the end

of the transit as seen by the observers A and B o n the Earth by

∆β

' ω

∆t

, ∆β

' ω

∆t

(3.3)

Now applying Pythagoras theorem to the triangles in Fig. we write



= [d

E−S

]

+ [d

E−S

(∆β

)]

= [d

E−S

]

+ [d

E−S

(ω

∆t

)]

(3.4)



= [d

E−S

]

+ [d

E−S

(∆β

)]

= [d

E−S

]

+ [d

E−S

(ω

∆t

)]

(3.5)

from which it follows

(α

− α

) =

∆



−



∆t



−

∆



−



∆t



(3.6)

The angle θ



can be expressed as



E−S



E−V

E−S



= 0.28 θ

(3.7)

After substituting the above expression of θ



and the angular separation into

Eq (3.2), the equation for Earth-Sun distance given in Eq (3.1) in reads

E−S



0.72

0.28







∆



−



∆t



−

∆



−



∆t



(ω

∆t

)

− (ω

∆t

)





(3.8)

In this formula the only unknowns are the durations of the transits, which can

be measured, and hence the distance to the Sun can be calculated. Note that

in the above expression we have ignored the eﬀect of the rotation of the Earth

and assumed that the orbits of Venus and Earth were circular. Taking into

account these corrections will improve the accuracy of the calculated value of

In 1919, Kepler established a relation between the square a planet’s period and its average

distance to the Sun, also known as Kepler’s third law of planetary motion. Consequently, the

distances of the known planets at that time (Uranus and Neptune were not discovered yet) to the

Sun in units of the Earth-Sun distance were known.

This is because the ratio of the size the Earth to the distance to Venus is extremely small.

– 10 –

the astronomical unit.

In 1771, the French astronomer Jerome Lalande used the combined the

data of the transit that occurred in the years 1761 and 1769 and calculated the

distance to the Sun, and hence the astronomical unit, AU, to be

]

(Transit)

= 153 million kilometers ± 1 million km (3.9)

which is just 2% higher than the actual value.

• More precise measurement of Earth-Sun distance

An accurate determination of the Earth-Sun distance has been obtained using

radar echoes where radio waves are transmitted toward an astronomical body,

such as a planet

. By measuring the time it takes the radio pulses to come

Figure 5: Using radar echoes to determine the Earth-Venus distance at the closest

approach, and hence the determining the Earth-Sun distance.

back we can determine the distance to that planet. The ﬁrst attempt to apply

this technique was in 1958 by the MIT Lincoln Laboratory where they directed

radar waves toward Venus, however the results were ambiguous. Three years

later, NASA’s Jet Propulsion Laboratory (JPL) directed a beam of radio waves

towards Venus and, for the ﬁrst time in history, they received the radar echoes

One cannot use radar ranging to measure the distance to the Sun directly because radio signals

are absorbed at the solar surface and not reﬂected to Earth.

– 11 –

after few minutes

. So by measuring the time it take the reﬂected radar pulses

oﬀ Venus when it is closest to Earth

and multiplying it by the speed of light

we obtain twice its distance to Earth at closest approach. For instance, it was

found that when Venus is closest to Earth, it took about 280 seconds to get

the radar echoes. Multiplying this round-trip travel time of the radar signal by

the speed of light (300,000 km/s), we obtain twice the distance from Earth to

Venus. Thus

(1 − 0.72) × d

E-S



3 × 10

km/s × 280 s



= 42 ×10

km (3.10)

from which it follows

E-S

' 150 × 10

km (3.11)

4 What is the average Earth -Moon distance ?

Quick Answer

E−M

= 384, 400 km

More Details

• Accurate Measurement of Earth-Moon distance

The Apollo missions 11 (1969), 14 (1971), and 15 (1972) deployed a retro-

reﬂector arrays on the Moon to measure the distance between surfaces of Earth

and the Moon from the duration that light takes from a very high-powered laser

pulses directed to the mirrors on the Moon to bounce oﬀ. It take light reﬂected

oﬀ the mirror anywhere between 2.34 to 2.71 seconds, depending how is the

Moon at the time of the measurement. Thus, the distance is in the range

E−M

∈ [351, 000 − 406, 000 km] (4.1)

This experiment, called the Lunar Laser Ranging (LLR), can measure the

distance between Earth-and Moon with an impressive accuracy of a millimeter.

After six minutes and half.

The same method can also be applied when the Venus is at the maximum elongation.

The speed of light is 300, 000 km/s.

– 12 –

• The gradual recession of the Moon

The Moon is gradually recessing from Earth of 3 cm/year.....

due to tidal eﬀect...... which will be discussed later

5 How did the ancients estimate the size of the Earth?

Quick Answer

Around 240 B.C, Eratosthenes

of Cyrene

determined the circumference of

the Earth by measuring the diﬀerence in the angles between the incident light

and the vertical at two diﬀerent locations, and using simple geometry.

He was a proliﬁc Greek scholar; he was a poet, an astronomer, geographer, philosopher, and

mathematician. He was also the chief librarian of the great library of Alexandria.

It’s a city in modern day Libya.

More Details

• Eratosthenes’ estimate of the size of the Earth [240 BC]

One day, while reading a papyrus book

in the great library of Alexandria,

Eratosthenes [276 −194 BC], came across the statement that at noon during

the summer solstice

, the light shone straight down into the bottom of a very

deep well

in the Egyptian city of Syene (modern-day Aswan)

, which means

that the Sun is directly overhead. At the same time in Alexandria (where he

lived), almost due north of Syene, he observed that the Sun light casts a shadow.

By measuring the length of the shadow casted by a tower in Alexandria, he

determined the angle between the vertical tower and the Sun’s ray

to be

equal to 7.2 degrees. He also knew that the distance between Alexandria and

Syene is 5000 Stadia, where a stadion

repesents the unit of lengths used at

that time. Unfortunately, there is not an exact conversion of "1 stadion" to

our unit metric system, however, many historians put it in a range between

0.16 km and 0.18 km. So, Eratosthenes reasoned that if 7.2

◦

decomposes the

Papyrus is a thick paper-like material that was used in ancient times as a writing surface.

This is when the Sun reaches its highest position in the sky, which occurs between June 20

and June 22 in the Northern Hemisphere, and between December 20 and December 22 in the

Southern Hemisphere.

The water was illuminated but not the walls of the well.

It is close to the tropic of cancer (23

◦

.5 N).

Knowing the height of the tower and the length of the shadow one can deduce the angle between

the vertical tower and the Sun’s ray.

In Latin it is "stadium".

– 13 –

Figure 6: Measuring the size of the Earth.

circumference of the Earth, C, to an arc of length 5000 stadia, then

C =

360

◦

7.2

◦

× 5000 Stadia = 250, 000 Stadia ∈ [40, 000 − 45, 000 km] (5.1)

Thus, the radius of the Earth is

⊕

[40, 000 − 45, 000 km]

2π

∈ [6370 − 7325 km]

This is within about 8% accuracy of the true value, which is quite remarkable

considering the very simple method that he used for this measurement

This result is indeed remarkable given that it was based on some erroneous assumptions. For

instance, it was assumed that Syene is due south of Alexandria; whereas it is actually 3

◦

east of it.

Also, at the summer solstice the Sun light is not precisely overhead at Syene, but about 0.4

◦

from

the vertical. Another inaccurate quantity is the angle that a shadow casts on the day of summer

solstice at Alexandria; the true value is 7.76

◦

instead of 7.2

◦

that Eratosthenes used. It turns out

that these errors almost cancel each out, and the estimate of the circumference would lie within the

the range given in (5.1) when using the unit of kilometer.

One might wonder what if Eratosthenes assumed that Earth is ﬂat. In this case, for the tower

to cast a shadow the Sun has to be be very close to the Earth Surface. With an angle of 7

◦

, the

distance from the Earth to the Sun would be

5000 stadia

tan 7

◦

∈ [6370 − 7325 km]

which is of course unacceptable since at that time it was already known (Aristarchus; 270 B.C)

that the d

was about 20 times the distance from Earth to the Moon (which we know now that

it is smaller by a factor of about 20 than the actual value).

– 14 –

• Al Biruni’s

method for measuring Earth’s radius [11

A.D]

6 What is the simplest clear experimental evidence for Earth rotation ?

Quick Answer

In 1851, the French physicist Leon Foucault, hang up a very long pendulum

from a ceiling at the Pantheon in Paris and let it swing, and showed that its

path "appeared " to be slowly shifting, which was a clear evidence that the

Earth rotates around its axis.

More Details

Figure 7: A schematic of the Foucault pendulum.

Abu Raihan Al-Biruni (973 − 1048) was a famous astronomer, physicist and geographer.

– 15 –

• To understand how does Foucault’s pendulum demonstrates the rotation of the

Earth around its axis, we need to discuss the so called Coriolis force, which

was named in honor of the French mathematician Gustave Coriolis

who

proposed the existence of pseudo-force (i.e. an apparent force) that a body

experiences when it is moving in a rotating frame. To be more speciﬁc, let us

consider a body that is located at a point (r, θ, φ) in spherical coordinates. Its

velocity as seen in a reference frame, R, attached at the Earth’s surface (see

Fig. 7) is

υ|

= υ

r + υ

θ + υ

φ (6.1)

Here

θ, and

φ form an orthonormal basis in the spherical coordinate system.

Then, the Coriolis force is given by

coriolis

= −2mω ×υ|

(6.2)

where ω = ω

z is the angular velocity vector of the Earth. Using the vector

identities

z ×

r = sin θ

φ,

z ×

θ = cos θ

φ,

z ×

φ = −sin θ

r − cos θ

θ (6.3)

Then, Eq (6.4) reads

coriolis

= 2mω

sin θ

r + υ

cos θ

θ − (υ

sin θ + υ

cos θ)

(6.4)

For example, if a body is moving north, i.e. υ

= υ

= 0 and υ

< 0, the

Coriolis force acts eastward.

• Foucault suspended an iron ball of mass m = 28 kg with a very long wire

length l = 67 m from the dome of the Pantheon in Paris. Due to the rotation

of the Earth, the plan of oscillations will rotate with respect to the surface

beneath it. From the view point of an observer in the inertial frame, there are

just two forces acting on the bob, the tension T in the string and the force

of gravity mg, whereas in the rotating frame of the earth, there are also the

centrifugal and Coriolis forces. However, the centrifugal force is a second order

in in ω, and so its eﬀect will be neglected. Thus, the equation of motion of the

mass m in the earth’s frame (R) is given by

r = mg + T −2m ω ×

r (6.5)

= −mg

z + T (

z cos θ −

x sin θ cos φ −

y sin θ sin φ) −2m ω ×

Coriolis lived during the 18th and 19th centuries, and published his work on the force caused

by the rotation of the Earth in 1835.

Recall that the oscillation period of a pendulum in an inertial frame is ∝

l/g, and hence with

very long wire it takes the pendulum longer time to swing back and forth. Making the pendulum

very heavy surpasses the eﬀect of air resistance.

– 16 –

For small oscillations, we can consider θ ' 0, and in this case the component

of the tension of the string along the z-axis cancels the weight. Then, in terms

of components, Eq (6.5) reads

¨x − 2ω

˙y + ω

x = 0, ¨y − 2ω

˙x + ω

y = 0 (6.6)

where ω

= ω sin ψ, and ω

= g/L is the natural frequency of simple pendulum.

Choosing the initial conditions x(0) = x

, y(0) = 0, ˙x(0) = 0, and ˙y(0) = 0,

the solution to the above system of diﬀerential equations is given by

x(t) =

(ω

sin ω

t sin ω

t + ω

cos ω

t cos ω

t) ' x

cos ω

t cos ω

y(t) =

(ω

cos ω

t sin ω

t − ω

sin ω

t cos ω

t) ' −x

sin ω

t cos ω

The above solution shows that due to the Coriolis force, from the point of view

of an observer on the ground it looks as the pendulum’s plane of oscillations

precess clockwise at constant angular velocity Ω = −ω

= −ω sin ψ with

respect to the Earth. However, from a view point of an observer in a reference

frame not attached to the Earth the plane of oscillation stays ﬁxed and it is

the Earth which rotates beneath it.

• This eﬀect is very small when observed in short time interval, but as the time

goes by, the orientation of the oscillation plane changes more and more. Thus,

at latitude, ψ, at which the pendulum is placed, the time, T , it takes the

oscillation plane of the pendulum to make a full rotation is

T =

2π

ω|sin ψ|

⊕

|sin ψ|

(6.7)

where T

⊕

is a sidereal period of the Earth, i.e. with respect to ﬁxed stars. For

instance, at the Panthéon in Paris, where the latitude is ψ ' 49

◦

, the period

of the oscillation is

T |

Paris

' 31 hrs 48

(6.8)

• It is worth mentioning that around 1650, almost two centuries before Foucault,

the astronomer Giovanni Battista, suggested that if the Earth is rotating

around its axis, then by dropping a ball from a a tower of about 75

meters tall it would land about one centimeter to the side. However, at

the time it was not easy to make a measurement with such precision to be able

to prove the Earth’s rotation around its axis.

Here x = l sin θ cos φ and y = l sin θ sin φ.

– 17 –

7 How fast does Earth revolve around the Sun?

Quick Answer

(⊕)



Equator

= 1675 km/hr ' 0.47 km/s

More Details

• The time it takes the Earth to complete one full rotation around its axis is

23 hours and 56 minutes. Note this is the sidereal day, and it is diﬀerent

from the 24 hours solar day. The former corresponds to the time that a

point on the surface of the Earth return to the same point with respect to to

some background stars, whereas the solar day is the time it takes the Sun to

return to the same point in the sky. However, since the Earth is orbiting the

Sun, it takes 4 minutes for the Sun to return to the same point. Thus, knowing

that the Earth’s circumference at the equator is 40075 km (see A 2), the speed

with which a point on its surface at this latitude goes in a full circle around

rotational axis of the Earth is given by

(⊕)

40075 km

(23 + 0.93) hours

' 1675 km/hr

• Because the Earth is spinning, we (and every object on it) are subject to the

centrifugal force that tries to push us oﬀ into space. However this centrifugal

force is just 0.3% the force of gravity, and hence, its eﬀect is negligible.

• We do not feel the diﬀerence in speed either way because everything around us

is moving at the same relative speed, even the atmosphere.

• If for some reason the Earth suddenly stopped rotating, we and every thing on

the Earth will ﬂy oﬀ with speed of 1675 km/hr, but not large enough for us be

launched into space

. However, the atmosphere would continue moving at the

original speed of the Earth’s rotation, producing large wind currents.

There are other dramatic eﬀects that make life on Earth unbearable, for example, half the

Earth will be permanently facing heat of the Sun, while the other half be exposed to the cold of

the space.

– 18 –

8 What is the relationship between sidereal and synodic periods ?

Quick Answer

• Synodic period of a planet

• Formula [Derived by Copernicus]











−

where E is the sidereal period of the Earth, P sidereal period of the planet,

and S its synodic period.

More Details

Derivation of the relation between synodic and sidereal periods

First, we note that Earth and any other planet move in their orbits at the rates

Earth

= 360

◦

/E, ω

planet

= 360

◦

/P (8.1)

Let us chose t = 0 to be the instant when a superior planet is in opposition. Then,

at the next opposition, the Earth has completed one orbit and then traverse an

additional angle ∆θ

Earth

= S ×ω

planet

during the time interval δt = (S − E), i.e.

Earth

(S − E) = ω

planet

(8.2)

or, equivalently,

−

[superior planets] (8.3)

For inferior planet, we just make the interchange E ↔ P in the above formula, and

obntain

−

[inferior planets] (8.4)

– 19 –

Planet P [Yrs] S [Yrs; calculated] S [Yrs; observed]

Mercury 0.24 0.3175 0.3175

Venus 0.62 1.604 1.599

Earth 1.00 (365.25 days) —– —–

Mars 1.88 2.111 2.135

Jupiter 11.86 1.092 1.092

Saturn 29.46 1.035 1.035

Uranus 84.01 1.012 1.012

Neptune 164.80 1.006 1.006

9 What is the Copernican model of the solar system ?

Quick Answer

• In 1543, Nicolaus Copernicus

, proposed that the Sun was stationary

in the center of the universe and the Earth and all the known planets

revolved around it in circular orbits.

• The Earth rotates around its axis, producing the motion of the ce-

lestial objects in the sky.

• This is known as the Heliocentric model of the solar system.

• At the time of Copernicus the only the known planets were: Mer-

cury, Venus, Mars, Jupiter, and Saturn.

He was a Polish mathematician, astronomer, and a priest.

More Details

• Some Historical Background

– Aristarchus of Samos [circa 300 B.C] was the ﬁrst to propose the

heliocentric model for the solar system, where the Sun is the center

of the solar system with the Earth orbiting it. In addition, he suggested

that the Earth rotates on its axis, causing day and night and accounting

for the daily motion of the stars. His reasoning was that it was unnatural

for the Sun to orbit the Earth which he estimated its size much smaller

than the Sun.

– His idea was rejected in favor of Aristotle’s geocentric model in which

the stationary Earth is the center of the universe

and the Sun,

According to Aristotelian worldview, if the Earth spins around its axis from west to east, we

should feel a wind blowing constantly from the east.

– 20 –

Moon, and all planets revolve around it

. One of the physical argument

that many Greek astronomers had against Aristarchus model is that if the

Earth was orbiting the Sun why no stellar parallax

has been observed

– However, the geocentric model could not account for the retrograde

motion, which is the occasional apparent westward motion of a planet

against the stars. Around 150 BC, Hipparchus of Rhodes (160 − 130

BC) developed a model, building on ideas of the great geometer Appolo-

nius of Perga who lived a century earlier

, to account for the retrograde

motion. He proposed that each planet moved with constant speed on

a small circle, called an epicycle, whose center moved uniformly on a

larger circle, called a deferent (see Fig..).

– In the 2nd century A.D, Claudius Ptolemy (100 − 170 AD), noted

that Hipparchus model did not ﬁt the data well as the observed speed

of the planets, Moon, and the Sun, change during diﬀerent stages

. For

that, he reﬁned the epicycle-diﬀerent mechanism by introducing the con-

cept of equant (see ﬁgure ..), a point on the diameter of the deferent but

at a position opposite to that of the Earth from the center of the defer-

ent

. So according to Ptolemy, epicycle still move about the center of

the deferent, but the uniform motion about the center of the deferent

is now replaced by the uniform angular motion about equant. For the

next fourteen centuries, this model remained the accepted description of

the planetary motion in the solar system.

– Although the the Ptolemaic geocentric system was successful in pre-

dicting explaining the planetary motion, the epicycles, eccentrics, and

equants that each planet requires to ﬁt the observational data made the

Eudoxus (circa. 400 BC), one of Plato’s students, was the ﬁrst to propose a universe where all

objects in the sky sit on moving spheres, with the Earth stationary at the centre. This is known as

"the homocentric" model of the universe. He needed a set of 27 interlocking spheres to explain

the observed motion of the celestial bodies: the exterior sphere carries the ﬁxed stars; each planet

requires four spheres (ﬁve planets were known at that time); the Sun and the Moon require three

spheres each.

For the deﬁnition of stellar parallax see question 22.

Of course, this argument turns out to be incorrect after the invention of the telescope, where

the parallax of many astronomical objects were measured.

He was Greek geometer and astronomer who lived during circa 247 222BC, known for his

Conics, a treatise in eight books where he introduced and named the conic sections, namely the

ellipse, parabola, and hyperbola.

Of course, Hipparchus was aware of this problem, in particular for the Sun and the Moon. To

solve this issue he proposed put the Earth not at the center of the deferent, but oﬀ its center by a

small amount (i.e. eccentric), which he estimated to be about 1/25 the radius of the Sun’s deferent.

So, as in the Hipparchus model, the Earth is not at the center of the deferent.

– 21 –

model incredibly complicated. This lead Nicolaus Copernicus in 1543

to propose a Sun-centered solar system. In the heliocentric model of

Copernicus the apparent retrograde can be easily explained as due to the

fact that the earth orbits the sun faster than the outer planets (i.e. Mars,

Jupiter, and Saturn) so that the position of one of such planets appear

to be moving backward when the Earth overtakes it

. Moreover, in the

Copernican model the rotation of the Earth’s rotation around its axis

accounts for the apparent daily rotation of the stars.

– In 1576, Tycho Brahe (1546 −1601), a Danish noble man and as-

tronomer, set up a state-of-the art observatory to measures planetary

and stellar motions with unprecedented accuracy

. To help him with this

endeavor, he hired Johannes Kepler(1571 − 1630), a German mathe-

matician,...

– After Brahe died

, Kepler used the highly precise observations on the

planet Mars by Brahe

and tryied to determine the shape of the orbit

that best ﬁt the data. He found that data would ﬁt more accurately if

Mars move in an ellipse rather than a circle

• Copernicus Derivation Planet’s distance from the Sun

Before we show how Copernicus determined the distances from a planet to the

Sun using Euclide’s geometry, we need some deﬁnitions (see Fig. ??).

? Inferior planets are the planets whose orbits are closer to the Sun than Earth,

i.e. the planets Mercury and Venus.

? superior planet are those planets whose orbits are beyond that of the Earth.

? Elongation is the angle between the sun and a planet as seen by an observer on

earth , which for a superior planet it can vary from 0

◦

to 180

◦

, whereas for inferior

planets the maximum elongations are 28

◦

for Mercury and 47

◦

for Venus.

? Opposition is a conﬁguration when a superior planet and the Sun are on opposite

As seen from Earth, the two inner planets, Mercury and Venus, donâ€™t exhibit retrograde

motion because they move faster than Earth.

Why we don’t feel a wind .....?

Brahe built his observatory in an island near Copenhagen which was given to him by the king

of Denmark, Frederick II who was a great patron of science and art.

was long assumed to have died from poisoning. Recently, archaeologists pored over his skeleton

to reveal the real cause of his early death: a fatal combination of obesity, diabetes, and alcoholism

(see [5]).

This includes 10 observations at oppositions, which Kepler used to determine the correct shape

of Mars orbit.

Initially, Kepler succeeded in ﬁnidng circular orbit for Mars that ﬁtted all the observational

data to within 2 arcminutes accuracy.

– 22 –

sides of the Earth, and in this case the planet would arise when the Sun sets.

? Quadrature conﬁguration occurs when the elongation is 90

◦

, and hence an inferior

planet can never be in quadrature.

– Distance of an Inferior Planet to the Sun

In Fig. we show an inferior planet at its greatest elongation, which we denote

as θ

max

. It can be shown using simple geometry that the angle between the

line Earth-planet, EP, and the line Sun-planet, SP, form a 90

◦

angle. Thus, by

measuring θ

max

, one can determine the planet-Sun distance, d

(inf)

P S

, in term of

the Earth-Sun distance, d

= 1 AU, i.e.

(inf)

= sin θ

max

= sin θ

max

[AU] (9.1)

Using the measured values of θ

max

for Venus and Mercury, we obtain

Mercury−Sun

= 0.47 [AU]; d

Mercury−Sun

= 0.73 [AU] (9.2)

It is remarkable that the above estimate for Venus is in excellent agreement

with modern measurement of the distance from the Sun. For Mercury the

calculated value is larger by 20% than the current measurement, which is due

to the fact that the orbit of mercury is more elliptical

than the one Venus,

which means that the θ

max

is not strictly a constant. So, a better estimate of

Mercury-Sun distance can be obtained by by measuring the diﬀerent consec-

utive maximal elongations over one year, take the average, and use it in the

above for formula to compute the distance to the Sun. Observations give an

average angle θ

max

' 22

◦

, which yields

Mercury−Sun

' 0.38 [AU] (9.3)

which is in very good agreement with the modern value.

– Distance of a Superior Planet to the Sun

At quadrature, the distance from a superior planet to the Sun, d

(sup)

P S

, is the

hypotenuse of the right triangle ABC, and hence

(sup)

cos θ



cos θ



[AU] (9.4)

where θ is the angle between the planet and the Earth as viewed from the Sun

in this conﬁguration. Thus, by measuring the number of days, ∆T , it takes for

a superior planet to move from an opposition to the next quadrature, we can

express θ as

θ =

360

◦

∆T

−

360

◦

∆T

(9.5)

Mercury ’s orbit has eccentricity e =' 0.2, the largest of all the planets in the solar system.

– 23 –

Here E and P are the sidereal periods of the Earth and the superior planet in

days. Using the formula for the sidereal and synodic that we derived in the

previous question, the above equation reads

θ =

360

◦

∆T

(9.6)

with S is the synodicl period of the planet. Hence, the planet-Sun distance can

be determined in terms of the measured quantities ∆T and S, and we get

(sup)

cos



∆T

× 360

◦



−1

[AU] (9.7)

• Kepler’s Laws of Planetary Motion

(i) Law of Orbits [1609]:

The planets orbit the sun in ellipses, with the Sun at one focus.

(ii) Law of Areas [1609]:

The line joining the Sun and a planet sweeps equal areas in equal times.

(iii) Law of Periods [1619]:

The square of the orbital period of a planet is proportional to the

cube of the semi-major axis of the ellipse.

In 1687, Isaac Newton used his universal theory of gravitation to deriver Kepler’s

laws and also determined the constant of proportionality between the square of the

orbital period of a planet, [τ

(rev)

]

, and the the cube of the semi-major axis of the

ellipse, a

, as

[τ

(rev)

]

4π

G (M



+ m

)

4π



(9.8)

where M



and m

are the masses of the Sun and the planet, respectively. The above

formula is known as Kepler’s third law.

10 What is the modern view of the solar system?

Quick Answer

It consists of

• Sun

• 8 planets

• More than 100 planetary satellites

• 5 dwarf planets

• Asteroids, meteoroids, and comets

– 24 –

More Details

• Basic properties of the Sun

– It is 109 times bigger than Earth.

– It makes up 99.8% of the mass of the entire solar system.

– It is located at 26, 000 ly from the center of the Milky Way galaxy.

– The Sun, and the whole solar system, orbits the center of the Milky Way

galaxy once each 260 million years.

• The Planets

After Newton’s time, with the improvement of telescopes, Two more planets

were discovered: Uranus [1781]

and Neptune [1846]

. So in total there

are 8 planets (see table 1 for details), and they are divided into two types:

(a) Terrestrial Planets

= Mercury

, Venus, Earth, Mars. They

have solid surfaces and are mainly composed of silicate compounds, and

their average mass densities range between 3.5 and 5 g/cc.

(b) Jovian Planets

= Jupiter, Saturn, Uranus, Neptune. They

have no solid surfaces and their outer layers are composed of light gases.

The average densities of Jovian planets are in the range 0.7 to 1.6 g/cc.

• More Moons

Except Mercury and Venus all other planets have Moon. There are at least

175 known Moons orbiting six of the eight planets

. The 6 largest Moons in

our solar system are:

Ganymede [Jupiter], Titan [Saturn], Callisto [Jupiter], Io [Jupiter], Earth

Moon, and Europa [Jupiter].

• Asteroid Belt

These are rocky interplanetary bodies that are remnants of the planets forma-

tion, which orbit the Sun at a distance [2.2 − 3.2 AU], forming a belt of

It was discovered accidentally by William Herschel

Its existence was predicted by Leverier , a French mathematician, and discovered by....

The term "terrestrial" comes from the work "terra" which means land.

Note that although Mercury is closer to the Sun than Venus, the later is hotter. The reason for

this is that Mercury has no atmosphere and so its surface reﬂects most of the radiation and heat

into space, whereas Venus has ticker atmosphere which consisting mainly of carbon dioxide (see

table 5), a powerful greenhouse gas, which trap the heat into the atmosphere and make it hotter.

The term "jovian" is derived from Jupiter, and is associated with Jupiter-like planets.

For example, Jupiter has more than 60 Moons, three of them larger than the Earth’s Moon

– 25 –

Name <d

> [AU] e δ

inc

[

◦

] R/R

⊕

min

surface

max

surface

[C] P

surface

[atm] τ

(rev(

[Eyears]

Mercury 0.39 0.206 7.0 0.38 - 170/+430 0 0.24

Venus 0.7 0.007 3.4 0.95 + 465 93 0.62

Earth 1 0.017 0.0 1 - 89/+ 58 1 1

Mars 1.5 0.093 1.9 0.53 - 143/+ 35 0.006 1.88

Jupiter 5 0.048 1.3 11.2 - 108 No-SS 11.86

Saturn 9.5 0.054 2.5 9.45 - 139 No-SS 29

Uranus 19 0.047 0.8 4.01 - 197 No-SS 84

Neptune 30 0.009 1.8 3.88 - 201 No-SS 164.8

Table 1: Some general properties of the planets : < d

> the average distance to

the Sun, e is the eccentricity of the elliptical orbit, δ

inc

is the planet’s orbit to the

plane of Earth’s orbit, R/R

oplus

the ratio of the radius of a planet to that of the Earth,

surface

surface temperature, P

surface

the atmospheric pressure at the surface, and τ

rev

the period of revolution around the Sun in units of Earth years. The abrevition "NO-

SS" stands for "No Solid Surface". Note the jovian planets have no solid surface the

measurements of the temperature are taken from a level in the atmosphere equal in

pressure to sea level on Earth.

about 1 AU lying between the orbits of the planets Mars and Jupiter,

and there are about two millions asteroids in belt, about a million of them

have more than a kilometer wide

. The ﬁrst and largest asteroid in this

belt was in 1801 by the Italian astronomer Giuseppe Piazzi, which he named

it Ceres after the Roman goddess of agriculture and fertility. Sometimes aster-

oids collide with each other and break up into small objects called meteoroids.

Their size ranges. from dust grains to about few meters wide objects. When a

meteoroid enters the atmosphere .....

• Dwarf planets

There are 5 dwarf planets in the solar system: Pluto, Ceres, Pluto, Make-

make, Haumea, and Eris.

• Comets

They are made of rocks, dust and ice; sometimes it is called dirty snowball,

which go around the Sun in elliptical highly-eccentric orbits. As a comet gets

closer to the Sun, the Solar radiation vaporizes some of the comet’s surface

material causing them to form a tail of dust and ionized gas. Comets originate

in the Kuiper Belt at 30 − 50 AU and the Oort Cloud at about 50, 000 AU ,

Based on geological evidence, around 66 million years ago Earth was hit by an asteroid of 10 km

in diameter at a speed of 10 km/s that caused the extinction of all non-avian dinosaurs and about

70% of all plant and animal species.

– 26 –

Figure 8: The Sun, the eight planets, and the ﬁve dwarf planets in the solar system

[credit: NASA].

Name <d

> [AU] R/R

Mercury

average

surface

[C] Year of discovered # of Moons

Ceres 2.8 R - 106 1801 0

Pluto 0.7 R - 228 1930 5

Haumea 1 1 - 240 2004 2

Makemake 1.5 R - 243 2005 1

Eris 5 R [- 248 to - 232] 2005 1

Table 2: Some general properties of the 5 dwarf planet in our solar system and the

years they were discovered.

orbiting the Sun in elliptical highly-eccentric orbits. Comets are usually named

after their discoverers. For example, the famous Halley’s comet can be visible

from Earth every 75 years. The last time it was seen was in 1986, thus it is

expected to return in 2061.

11 What is Bode’s Law?

Quick Answer

• In 1772, the German astronomer Bode proposed an empirical rule that

gives approximate distances of planets from the Sun.

More Details

– 27 –

Start with the following sequence of numbers

0, 3, 6, 12, 24, 48, 96, 192, 384

Now, add 4 to each number above:

4, 7, 10, 16, 28, 52, 100, 196, 388

Then, divide each number by 10:

0.4, 0.7, 1.0, 1.6, 2.8, 5.2, 10.0, 19.6, 38.8

If we set the distance Earth-Sun = 1 (AU), then we see that the numbers in the

above sequence approximately represent the distance of the planets from the Sun,

except the the value 2.8. However, in 1801, an astronomical object at about 2.8 AU

from the Sun was discovered. It can be expressed in a mathematical formula as

= 0.4 + 0.3 ×2

, n ∈ 0, 1, 2, 3, 4, 5, 6 (11.1)

Here the number 0.4 in the above relation represent the distance Mercury-Sun, n the

number of planet of the planet with n = 0 for Venus, n = 1 for the Earth, and so on,

and d

is the distance of the n

planet from the Sun.

0 1 2 3 4 5 6

Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune

Bode’s law 0.4 0.7 1.0 1.5 2.8 5.2 9.6 19.2

Actual distance 0.4 0.7 1.0 1.6 2.8 5.2 10 19.6

12 What is the universal law of gravitation ?

= G

where

, m

: masses of the two objects

G : Gravitational Constant

d: the distance between the two masses

More Details

– 28 –

• In 1687, Isaac Newton published the universal law of gravitation and his

three laws of motion in "The Mathematical Principles of Natural Philosophy",

commonly known as the Principia.

• In 1749, the French geophysicist Pierre Bouguer tried to measure the density

of the Earth using the deﬂection of a plumb line due to the attraction of a

mountain. That would allowed him to determine the strength of the universal

gravitational force. Unfortunately, he did not succeed as the deviations of

pendulum were extremely tiny to be measured. The same method was applied

by N. Maskelyne and C. Hutton in 1755 and determined that the average

density of the Earth to be between 4.5 − 5 g/cc, which is of the same order of

the current measured value.

• It was Henri Cavendish who, in 1798, performed an extremely delicate ex-

periment that allowed a precise measurement of the gravitational constant

The setup is as follows: A dumbbell consisting of two identical masses m

attached to its ends and suspended horizontally from a very thin torsion wire.

Two other much heavier balls of identical mass M are placed at the positions

shown in the ﬁgure(see Fig. 9).

Theses masses produce attractive forces on

the two masses m and cause the dumbbell to twist. The dumbbell will oscillate

back and forth before ﬁnally settling down at some tiny angle θ

∗

away from

the initial position due to the restoring torque provided by the wire. With θ

∗

being very small, the torque on the dumbbell that arises from the twist takes

the form of Hooke’s law

τ = −κθ

∗

(12.1)

Here κ is a constant that depends on the thickness and the material from which

the wire is made of. If we denote by d

∗

the separation at equilibrium between

the centers of the masses in each pair, then gravitational force between each

pair of is

= G

∗

(12.2)

Hence, the torque on the dumbbell due to the two gravitational forces is

(GMm/d

∗

) L, where L is distance between the small balls of the dumbbell.

Applying the equilibrium condition, i.e. the total torque equals to zero, we get

G =

κ θ

∗

MmL

(12.3)

Actually, Cavendish intended this experiment to be a measurement of the Earth’s density

relative to water through the precise knowledge of the gravitational interaction.

The experiment was originally devised by the geologist John Mitchell around 1780, but he died

before he could begin taking measurements. However, Cavendish substantially improved Mitchell’s

apparatus.

– 29 –

Figure 9: Cavendish’ experimental apparatus taken from [6].

Note that for θ

∗

large enough for it to be measured, the torsion proportionality

constant κ needs to be as small as possible, and so it is desirable to have l as

large as possible

. For that, Cavendish made use of high precision vernier scale

and a small telescope to aid the detection of the rotation angle, and measures

the displacement ∆S

∗

of the small ball

. So by measuring the The constant

κ can be obtained in terms the moment of inertia of the dumbbell, I, and its

oscillation period, T

, as follows. By twisting the dumbbell (in the absence

of the large masses) by small angle and let it oscillate around its equilibrium

position, the equation of motion reads

θ + κ θ = 0 (12.4)

This is just the equation of harmonic oscillator with frequency ω =

κ/I, and

so by measuring the period of the oscillations one can determine the constant

For a given material, the torsion constant is proportional to the fourth power of its diameter

and inversely proportional to its length.

It is a very challenging task to precisely determine the deﬂection angle even with the tele-

scope that Cavendish used. In modern experiments, to overcome this diﬃculty a small mirror

is attached to the center of the rod to reﬂect a light beam on a wall where the angular mo-

tion can be more easily observed. If the wall is at a distance D from the mirror, then when

the mirror rotates with an angle θ

∗

, the spot on the wall from the reﬂected light beam moves

a distance x = D tan 2θ

∗

' 2Dθ

∗

. A nice animation of this type of setup can be viewed at

https://www.youtube.com/watch?v=EE9TMwXnx-s.

– 30 –

κ from the relation

κ = I



2π



(12.5)

Plugging this expression of κ into (12.3) yields

G =

4π

∗

MmLT

(12.6)

or, equivalently, in terms of the displacement ∆S

∗

, we obtain

G =

8π

∗

∆S

∗

MmL

(12.7)

In this experiment Cavendish used m = 0.73 kg, M = 158 kg, d

:= (d

∗

+ ∆S

∗

) =

22.5 cm, L = 186 cm, and measured the oscillation period T

= 14.6 min and

the displacement of the small ball ∆S

∗

= 4.1 mm

. With these values one

ﬁnds

(Cavendish)

= 6.74 ×10

−11

N.m

/kg

(12.8)

which is very close to present-day measured value.

13 What is the minimum period of rotation a spherical astronomical object

can have ?

Quick Answer

A spherical object with mass density ρ can not have

a rotation period less than

min

11 g/cm

hours

If we approximate the moment of inertia by I = m (L/2)

+ m (L/2)

= ml

/2, then the

expression of G

reads

G =

2π

Iθ

∗

which is independent on the value of the mass m.

He determined the displacement with to an accuracy of better than 0.25 mm.

We should point out that Cavendish did not explicitly derive the expression of G, and in fact it

is not even mentioned in his paper. Instead he computed the density of the Earth which is related

to the gravitational constant (see A7). The ﬁrst reference to the universal gravitational constant

was in 1873 by Alfred Cornu and Baptistin Baile (which was denoted by the letter f), about 75

years after Cavendish’s work, and the modern notation with the letter G was introduced in 1894

Charles . V. Boys (see ref. [7]).

– 31 –

More Details

Let R be the radius of a spherical astronomical object with mass density ρ that

we assume to be uniform though out its volume. When the object is spining it is

subject to a centrifugal force, acting outward. For a point mass δm at the equator

of the object, this force is radial of magnitude F

= δma

, where a

is the centrifugal

acceleration, given by

(13.1)

In addition, gravity exerts a radially inward force F

= δmg, with g being the

gravitational acceleration at the surface of the object. It reads

g =

4π

GρR (13.2)

Here M is the mass of the object, and G is Newton’s gavitational constant (see sec-

tion 12). In obtaining the second equality we made use of the assumption that ρ is

constant.

In order for the spinning object not to disintegrate, the gravitational force must

be larger or equal to the centrifugal force, i.e.

υ ≤ R

4π

Gρ (13.3)

If T is the period of rotation, then we can use the fact that υT = 2πR to express

the above inequality as can we can

T ≥

3π

Gρ

=⇒ T

min

3π

Gρ

(13.4)

From this we that the minimum period depends only on the mass density of the

object and not its mass or radius. We also note that the denser the object is the

larger its rotation frequency is.

In table z, we show the minimum period required for a spining terrestrial planet,

a star similar to the Sun, and a neutron star. We also show the typical observed

values of the their rotation periods.

14 What is the mass of the Earth ?

Quick Answer

– 32 –

⊕

' 6 × 10

More Details

• Consider the Earth to be a perfect sphere of radius R

⊕

, and mass M

⊕

. Let

a test particle of mass, m, free falls toward the ground from some distance

much smaller than the Earth’s radius. Then, according to the universal law of

gravitation (see A6) and Newton?s second law of motion we have

⊕

+ r)

= mg(h) (14.1)

where h is the distance from the ground to the mass m at some instant t during

the free fall, g(h) is its acceleration due to gravity at the height h, and G is the

universal constant of gravity. After canceling the common factor m from both

sides of the equation, and neglecting the contribution of h as compared to R

⊕

we obtain

⊕

(14.2)

Here g

⊕

= g(h = 0) is the acceleration of gravity at the surface of the Earth

Thus, using R

⊕

' 6370 km (see A2), G = 6.67 × 10

−11

N.m

.kg

−2

(see A5),

and g

⊕

' 9.8 m/s

, we obtain

⊕

6.37 × 10

× 9.8 m/s

6.67 × 10

−11

N.m

.kg

−2

' 6 × 10

• Since we now know the mass of the Earth and its radius, we can calculate its

average density from the formula

⊕

(4π/3) R

⊕

' 5.5 g/cm

' 5.5 × ρ

(14.3)

Since most of the rocks have density of about 3 g/cm

, we conclude that the

Earth?s interior must be more dense than typical crustal rocks in order for the

whole thing to average to about 5.5 g/cm

The value of g

⊕

can be easily measured by either recording the positions, x(t), that a test particle

moves at diﬀerent instances, t, during the free fall, or, by measuring the period of oscillations of a

simple pendulum of a given length since its period is a function of the acceleration of gravity at the

Earth’s surface.

– 33 –

• Another method: If we know the distance from the Earth to the Moon, we

can determine Earth’s mass using Kepler’s third law and the measured value of

the gravitational constant. In the same way, one can infer the mass of a stars

that form a orbiting binary star-system with another one (see A12 for details).

15 What is gravity at the surface of an astronomical object ?

Quick Answer

(surface)

More Details

• The above expression for the gravitational acceleration can be obtained using

Newton’s universal law of gravitation and Newton’s second law of motion,

which for a test mass m located at (or near) the surface of an astronomical

object of mass M, reads

−

GMm

r = ma (15.1)

where

r is a unit vector directed from the center of the celestial body to the

test mass. Hence, the acceleration of gravity at the Earth’s surface is given by

a = −

r ≡ −g

(surface)

r (15.2)

with g

(surface)

is the magnitude of the gravitational ﬁeld at the Eearth’s surface.

• In table 3, we give the values of the acceleration of gravity for the Moon and

the planets in our solar system.

– 34 –

Object g [m/s] υ

(km/s)

Sun 273.4 617.5

Mercury 3.7 4.3

Venus 8.9 10.3

Earth 9.8 11.2

Mars 3.7 5

Jupiter 24.8 59.6

Saturn 9 35.6

Uranus 8.7 21.3

Neptune 11 23.8

Moon 1.6 2.4

Table 3: The values of the acceleration of gravity and the escape velocity for the

Sun, the Moon, and the Planets.

16 What is the escape velocity from the Earth?

Quick Answer

(⊕)

esc

= 11.2 km/s

More Details

• The escape velocity is the minimum speed υ

that an object needs to have

to break free from the surface of a large body such as moon, planet or a star.

Suppose that an object of mass m leaves the surface of stationary object of mass

M >> m, with initial speed υ

. If, for simplicity, we consider the astronomical

object to be spherical of radius R, and assume it has no atmosphere, then the

energy conservation E(r = R) = E(r), ∀r > R, implies that

mυ

− G

mυ

∞

(16.1)

where υ

∞

is the speed of the particle at r = ∞, i.e. at the point where the

mass m has completely escaped the eﬀect of the gravity due to the mass M.

Since the right hand side of the above equation can not be negative, the escape

velocity correspond to the value of υ

for which υ

∞

= 0, i.e.

esc

(16.2)

– 35 –

Figure 10: Escape velocity.

which is independent of the mass of the escaping object

. Applying the above

formula for the escape velocity to the Earth, and using M

⊕

' 6 × 10

kg (see

A6), we obtain

(⊕)

esc

= 6.9 m/s = 11.2 km/s (16.3)

• In table 3, we give the values of the escape velocity for diﬀerent planets in the

solar system and some other astronomical objects

• In the absence of the atmosphere, to get a spacecraft from Earth’s surface into

Earth’s orbit would require that the speed of the rocket to be 7.8 km/s

17 What is the age of the Earth?

Quick Answer

It is important to note that the above calculation neglects the eﬀect of air resistance.

Titan is the Saturn’s largest moon, and Europa is one of Jupiter’s moon.

Because o the Earth’s atmosphere, the rocket needs to have a speed of about 10 km/s to

overcome the air resistance.

– 36 –

Radioactive Element Symbol Half-life

Radon-222

222

Rn 3.8 days

Carbon-14

C 5, 730 years

Uranium-235

235

U 0.7 billion years

Potassium-40

K 1.25 billion years

Uranium-238

238

U 4.47 billion years

Thorium-232

232

Th 14.1 billion years

Rubidium-87

Rb 47 billion years

Table 4: Common radioactive elements (found in rocks, water, and air) .

⊕

' 4.5 billion years

• The oldest known rocks on Earth were dated at 4.28 billion years. They were

discovered in 2008, in northern Quebec, Canada.

• The oldest minerals found on Earth are zircon with an age of 4.40 billion years.

They were discovered in 2014 in the Jack Hills of western Australia.

More Details

• Radiometric dating:

The dating of a material, such as a rock or a fossil, is based on the method

of radiometric dating, which is the measurement of the radioactive decay

of an unstable isotope

(parent) into more stable element. In table 5 we list

some of the radioactive elements that are widely used in radiometric dating.

• The equation of a radioactive decay process:

Consider a system of radioactive nucleus, P, with N

(0) their number at the

instant t = 0. We assume that each particle decays into a stable daughter

nucleus, D, with equal probability per unit time, λ, also known as the decay

constant for the decay process. Then, during the time interval [t, t + dt], the

decrease in the number of the P particles, N

, and the growth in the number

of the D particle, N

, are given by

Radioactivity was discovered in 1896 by the French physicist Henri Becquerel.

Isotopes are atoms that their nuclei contain the same number of protons but have diﬀerent

number of neutrons. This means they have diﬀerent mass but their chemistry is unchanged since

the outer shell conﬁguration will be the same.

– 37 –

(

= −λN

= λN

(17.1)

Note that d (N

+ N

) = 0, which is implies that the total number of particles

(i.e. parents and daughters) is conserved, i.e.

+ N

= N

(0) + N

(0) (17.2)

Now the integration of (17.1) yields

(

(t) = N

(0) e

−λt

(t) = N

(0) + N

(t)



−λt

− 1



(17.3)

This exponential distribution of the parent particles can be interpreted in term

of a decay probability, dP(t), for a single particle to disintegrate between

the instant t and t + dt, given by

dP(t) = λe

−λt

dt (17.4)

Thus, the probability for particle to decay within a time t is

P(t) =

dP(t) dt = 1 − e

−λt

(17.5)

Using this probability distribution, the mean lifetime of the radioactive particle

P can be calculated as

τ =

t dP(t) = λ

−1

(17.6)

Another related quantity is the half-life, τ

1/2

, deﬁned as

1/2

' 0.7 τ (17.7)

In obtaining the equation of N

(t), we made use of Eq (??).

This is also known as Poisson distribution, and it is normalized to unity.

Equivalently, it can be obtained as

τ =

(0)

∞

t=0

t dN

(0)

∞

t=0

t dN

(0)

∞

t=0

t λN

dt = λ

∞

t=0

t e

−λt

dt = λ

−1

This relation can be shown as follows:

N(τ

1/2

) = N

−λτ

1/2

→ τ

1/2

ln 2

– 38 –

Now, suppose we know N

(0) or its ratio compared to some stable element in

the sample

, P

, then from Eq (17.3) the the age is simply

t = τ ln



(t = 0)

(t)



= τ ln



) (t = 0)

) (t)



(17.8)

We can generalize the above results to the case when the parent nucleus decay

to many daughter nuclei, D

, D

, ..., D

, with decay constants λ

, λ

, ..., λ

, i.e.

P → D

+ ...., i = 1, 2, ...., n

Then,











= −λN

dt,

= λ

dt, i = 1, 2, ..., n

d (N

i=1

) = 0

(17.9)

where λ =

i=1

is the total decay constant for these processes. Thus,

integrating the above diﬀerential equations gives











(t) = N

(0) e

−λt

(t) = N

(0) +

(0)



1 − e

−λt



, i = 1, 2, ..., n

i=1

= N

(0) +

i=1

(0)

(17.10)

• Carbon-14 dating [≤ 70, 000 years

The interaction of high energy cosmic rays (mostly protons, and helium nuclei)

with the atmosphere produce showers of many types of subatomic particles,

including energetic neutrons

. These neutrons collide with the nitrogen atoms

in the atmosphere producing a hydrogen and a carbon-14, i.e.

n +

N →

C + p

These

C atoms combine with oxygen molecules, forming carbon dioxide,

which spreads around at low altitudes, where it is used by the plants, and

hence by all living things.

At any particular time, the ratio of

C to

C is approximately 10

−12

in all

living organisms. We will denote this ratio by R

. However, once an organism

dies the carbon-14 starts to decay to nitrogen via the β-process:

C →

N + e

−

+ ¯ν

For example, one of the stable particle of the decay product.

Beyond an age of 50, 000 years the amount of

C left in them is so tiny that it can not be used

for a reliable dating.

This happens at high altitudes around 9 − 15 km in the atmosphere.

For thermal neutrons, σ(n +

N →

C + p) ' 1.8 × 10

−24

≡ 1.8 barn.

– 39 –

where ¯ν is the anti-neutrino, a neutral spin 1/2 particle with almost vanishing

mass. Thus, by measuring the ratio R(t) =

C(t)

, then compare it to the

assumed initial value R

in a living organism, scientists can determine the time

elapsed since it died. Using Eq (17.8) and the value of τ

1/2

C from table

5, we obtain

[t]

= 8022 ln



R(t)



years

• Potassium-Argon dating: [≥ 10

years]:

– Potassium is present in Earth’s crust

with a concentration of about

1.5%, and its naturally occurring radioisotopes,

K make up 0.12% of

natural potassium, and hence it can be used in dating rocks. About 11%

K decays to

Ar and 89% to

Ca via the following processes:

K →

Ca + e

−

+ ¯ν; [β process; 89%; λ

= 4.96 ×10

−10

years]

K + e

−

→

Ar + ν; [electron capture; 11%; λ

= 0.58 ×10

−10

years]

– However, because

Ca can be present both as radiogenic and non-radiogenic

and so only the ﬁrst decay process can not be used for dating rocks. In-

stead, geologists make use of the decay to argon by electron capture and

measure the

Ar/

K ratio in materials that trap argon. Furthermore,

since argon is a noble gas it can escape easily from the extremely hot

molten magma, and so it is reasonable to assume that there was no ar-

gon initially when the rocks formed. Thus, the argon in a rock today is

that which was produced by

K since the rock has cooled down.

– Now, applying Eq (17.10) to the decay processes above, we obtain

(t) =

(t)



λt

− 1



(17.11)

where λ = λ

+ λ

= 5.54 ×10

−10

years is the total decay constant of

to argon and calcium. Solving for t, we obtain

[t]

Ar/K



(t)





+ 1



= 1.8 ln



9.55



(t)



+ 1



billion years

As well in oceans and all organic materials.

It means can be produced by another radioactive decay process. So it would be impossible to

distinguish what proportion of

Ca that was produced by potassium.

– 40 –

– Uranium-lead dating: [≥ 10

years]

The naturally occurring uranium consists primarily of two radioisotopes:

238

U and

235

U, which produce

206

Pb and

207

Pb, respectively, with the

emission of α particles, via long decay chains. Another isotope of lead

204

Pb which is non-radiogenic, and so using Eq (17.3) one can get two

independent dates from the uranium-lead system:



206

204



(t) −



206

204



(0) =



238

204



(t)



238

− 1



(17.12)



207

204



(t) −



207

204



(0) =



235

204



(t)



235

− 1



(17.13)

Both these equations is has the form y = mx + y

, where y is the ratio of

the amount of lead produced from radioactive decay to

204

Pb, and they

are known as the isochron equations.Taking the ratio of the above

equations, we obtain

207

204

Now

−

207

204



207

204



Now

−



207

204





235

238



235

− 1

238

− 1



137.8



235

− 1

238

− 1

where we used the present day ratio

235

238

U is 137.8, as it has been

measured in terrestrial, lunar, and meteorite samples. The above equa-

tion is known as Pb-Pb isochrone equation. For a rock or meteorite,

this equation represents a straight line with [

207

Pb/

204

Pb]

Now

is, say on

the y-axis, and [

206

Pb/

204

Pb]

Now

on the x-axis, and the left-hand side of

the equation is the its slope. So if the initial ratio of P b are known or

inferred from minerals that essentially had no lead initially present in

them

, then one can determine their age.

18 What is the critical temperature of a planet’s atmosphere above which a

particular atmospheric component will be lost into space ?

Quick Answer

critical

GMm

More Details

Such as zircon, apatite, or monazite.

– 41 –

• The most probable speed of a molecule in a gas

Whether a planet has an atmosphere depends on the ratio of the molecular

speed in the gas and the escape velocity from the planet. Thus, we will ﬁrst

calculate the mean speed of molecules in a gas. For that, we recall that in

thermal equilibrium, the velocity distribution of particles of mass m is

f(υ)d

υ = N



2πk



3/2

−mυ

/2k

dυ sin θdθdφ (18.1)

Integrating over the angles θ and φ, we obtain the speed distribution function

f(υ)dυ = N





3/2

−mυ

/2k

dυ (18.2)

The most probable speed, υ

, is given by the speed at which the derivative of

f(υ) vanishes. We ﬁnd

−→ f(υ) =

3/2

exp



−



(18.3)

The speed υ

given above can also be expressed in terms of the molar mass M

of the gas molecules as

2RT

(18.4)

where R = 8.314 J/(gram-mol.K) is called the universal gas constant. Thus,

at a given temperature, the more massive the molecules of a gas, the lower

is their mean speed. This is why it is more easy for hydrogen and helium to

escape the Earth rather than does carbon dioxide, when the temperature of the

gas is large enough so that υ

> υ

esc

. With the Earth’s average temperature

of 15

◦

(i,e, approximately 280 K), the oxygen and nitrogen gases, have average

speeds of about 380 m/s and 406 m/s, respectively. These are much smaller

than the escape velocity from the Earth (see A(09)), which explains why these

gases are still present in the atmosphere.

• Naive estimate of the critical temperature

A particle can escape the surface of a planet if its kinetic energy exceeds the

gravitational energy of the planet. For a gas molecules, this is equivalent to

Note that the most probable speed is not the average speed. The later is given by

υ =

∞

υ f(υ)dυ =

= 2υ

– 42 –

the condition

rms

≥ υ

esc

2GM

(18.5)

which in terms of temperature, yields

T ≥

2GMm

(18.6)

Hence, for T ≤ 1000 K, a very tiny fraction of the molecules in the gas have

such high speed, and one might think this will have no eﬀect on particles loss

from the atmosphere.

In reality, even if υ

< υ

esc

, there will always be particles in the high-speed

tail of the distribution in (18.2) that have speed larger than the escape velocity

which leak into space. Although the fraction of such particles is tiny, the

remaining particles will redistribute them selves so that the high-speed tail get

reproduced so that their speeds follow a Maxwell-Boltzmann distribution, and

again the particles in the high-speed tail will escape into space. Therefore, a

more relevant quantity to estimate is the time scale for the gas molecules to

escape the planet’s atmosphere, as will be discussed below.

• Jeans thermal escape ﬂux

A particles can escape the planet’s atmosphere if the magnitude of the vertical

component of its velocity, υ

= υ cos θ, exceeds the escape velocity from the

planet, and does not collide with other particles in its motion upward. This

region where it is unlikely for a gas particle to collide with another one is

known as exosphere. If we denote by n

exo

the density of the gas particle in

the exobase

, then the escape rate is

esc

= n

exo

∞

esc

f(υ)υ

sin θdθ dυ dφ (18.7)

where



esc



(18.8)

After integrating (18.7) yields

esc

exo

√

(1 + Y

) e

−Y

(18.9)

This is the altitude above which it is unlikely for the particles to collide with each other.

– 43 –

The above expression is called Jeans thermal escape ﬂux. Note that the

smaller is Y

, the more signiﬁcant is the escape rate.

As an example, consider the gas molecules N

, which is the most common gas

in the Earth’s atmosphere. With the exobase at altitude of about 500 km at

which T ∼ 1000 K, we have

] = 0.78 km/s −→ Y

] ' 196 (18.10)

Thus, the escape rate of the nitrogen molecules is

esc

] ' 8 ×10

−56



exo

2.5 × 10

−3



−2

−1

(18.11)

We see that even if we the value n

exo

= 2.5 × 10

−3

, which is the num-

ber density of the molecular nitrogen gas at the sea level, the escape rate is

vanishingly small. To estimate the number of N

particles escaping from the

atmosphere per unit time, we multiply Φ

esc

by the surface area of the exobase,

4π (R

⊕

+ H)

, where R

⊕

is the Earth’s radius

. We get

dN[N

]

∼ −10

−41



exo

2.5 × 10

−3



Molecule

billion year

(18.12)

which shows that the Earth will hold its nitrogen (and oxygen) gas much longer

than the age of the universe.

• The time scale for particles to escape entirely

Let h be thickness of the region in the atmosphere where the probability for

particles to collide with each other is extremely small. Then, a rough estimate

of the time it takes for such particles to escape entirely from the atmosphere

into space is

esc

∼

esc

(18.13)

where n is the average number density of the gas above the exobase where

particles collision is very rare. Substituting the expression of τ

esc

in the above

equation gives

esc

∼ 3



exo



√

1 + Y



500 km



10 km/s

esc



minute (18.14)

In reality, at altitudes above 200 km the number density of nitrogen drops signiﬁcantly compared

at sea level.

– 44 –

Planet Components of the Atmosphere Comments

Mercury 42% O

, 29% Na, 22% H Extremely thin atmosphere

(Surface pressure ' 0)

Venus 96% CO

, 3.5% N

Runaway greenhouse eﬀect from CO

Earth 78% N

, 21% O

Thin atmosphere

Mars 95% CO

, 2.7% N

, 1.6% Ar Very thin atmosphere (Surface pressure ' 7 ×10

−3

atm)

Jupiter 90% H

, ∼ 10% He Height from its base > 5000km

Saturn 96% H

, ∼ 3% He Magnetic ﬁeld smaller than Jupiter

Uranus 83% H

, 15% He

, 2.5% CH

Blue-Green color of atmosphere from CH

Neptune 80% H

, 19% He

, ∼ 1% CH

Bright blue atmosphere from CH

Table 5: The gases in the atmospheres of the planets in our solar system.

Let us estimate for how long an Earth-like planet can hold an atmosphere of

hydrogen gas. We will assume that initially that the atmosphere contained a

substantial amount of hydrogen, and the temperature is about 1000 K at the

base of the exosphere

. Using Eq (18.4), we obtain

(H)

' 4 km/s (18.15)

Now, if we assume that hydrogen is abundant at altitudes between 200 km and

1000 km, i.e. h ' 800 km, we obtain

(H)

esc

⊕

∼ 12 days (18.16)

However, from observations, we know that hydrogen is one of the abundant

gases in the exosphere, which seems to be in contradiction with our estimate

above. This is because in obtaining τ

(H)

esc

⊕

we have assumed that there were

no source for hydrogen production in the atmosphere, which is not correct,

whereas at latitudes above 500 km, hydrogen is constantly produced from the

collision of high-energy photons from the Sun with water vapor and methane

molecules.

19 What type of electromagnetic spectrum reach the Earth’s surface ?

Quick Answer

Jeans wrongly assumed that the Earth’s upper atmosphere was very cold and concluded that

no hydrogen leaked into space.

The molecular mass of hydrogen is M = 1 g/mole.

– 45 –

• The range of the electromagnetic (EM) spectrum:

Type Wavelength Frequency range Energy

Gamma ray < 10 pm [> 30 EHz] > 10 keV

X ray 10 pm − 10 nm [30 EHz − 30 PHz] 10

eV − 10 keV

Ultraviolet 10 − 400 nm [30 PHz − 750 THz] 3 −10

Visible 400 −700 nm [750 − 430 THz] 2 − 3 eV

Infrared 700 nm − 1 mm [430 THz − 300 GHz] 0.1 − 2 eV

Microwave 1 mm −1 cm [300 GHz − 300 MHz] 100 µeV − 0.1 eV

Radio 1 cm − 100 km [300 MHz −3 KHz] < 100 µeV

• Transparency of the Earth atmosphere to EM radiation

The Gamma-rays, X-rays and UV with λ ≤ 300 nm get absorbed by

the ozone molecules in upper atmosphere, whereas most of the infrared

radiation are absorbed by the H

O and CO2 molecules.

More Details

• Example of typical astronomical sources of the EM spectrum

Type Typical Source Temperature

Gamma ray matter falling into black hole > 10

X ray gas in cluster of galaxies 10

− 10

Ultraviolet very hot stars 10

− 10

Visible stars; hot planets 10

− 10

Infrared very cool stars; planets; cool cloud of dust 10 − 10

Microwave/Radio cosmic microwave background ; electron moving in magnetic ﬁeld < 10 K

– 46 –

• Some major space-based telescopes across the EM spectrum

Name Space Agenc Cost[Billion] Launched-Termiated Wavelength range

COBE NASA 0.16 ($) 1989 - 1993 Microwave

HIPPARCOS ESA 0.60 (e) 1989 - 1993 Optical

HST NASA 10 ($) 1990 - UV, Optical, NIR

Chandra X-ray NASA 1.65 ($) 1999 - X-ray

XMM-Newton ESA 0.77 (e) 1999 - X-ray

WMAP NASA 0.15 ($) 2001 - 2010 Microwave

INTEGRAL ESA 0.30 (e) 2002 - Gamma-ray

Spitzer Space Telescope NASA 0.72 ($) 2003 - IR and Sub-mm

FGST (formerly: GLAST) NASA 0.69 ($) 2008 - Gamma-ray

Planck ESA 0.70 (e) 2009 - 2013 Microwave

Herschel Space Observatory ESA 1.10 (e) 2009 - 2013 IR and Sub-mm

GAIA ESA 0.65 (e) 2013 - Optical

• Some of the major ground-based telescopes across the EM spectrum

Name Dishes/Mirrors [in meter] Location Start-End Wavelength range

VLA 27 dishes ([25]; Y shape) New Mexico (USA) 1981 - Radio

Keck 2 Telescopes (I & II)/[10] Hawaii (USA) 1993 &1996 - Optical/NIR

VLT 4 Telescopes [8.2] Atacama desert (Chile) 1998 - Optical/NIR/MIR

HESS 5 Telescopes (1 [28] + 4 [12]) Namibia 2003 - Gamma-ray(indirect)

MAGIC 2 Telescopes [17] Canary Islands (Spain) 2004 & 2009 - Gamma-ray(indirect)

SALT 1 Primary Mirror [9.2] (South Africa) 2005 - Optical/NIR

VERITAS 4 Telescopes [12] Arizona (USA) 2007 - Gamma rays(indirect)

GTC 1 Primary Mirror [10.4] Canary Islands (Spain) 2007 - Optical/NIR

ALMA 66 dishes (54[12] + 12 [7]) Atacama desert (Chile) 2011 - NIR-Radio

FAST 1 dish [500] Guizhou (China) 2016 - Radio

• Acronyms

* KHz → Kilohertz = 10

Hz.

* MHz → Megahertz = 10

Hz.

* GHz → Gigahertz = 10

Hz.

* THz → Terahertz = 10

Hz.

* PHz → Petahertz = 10

Hz.

* EHz → Exahertz = 10

Hz.

* NIR → Near Infrared.

– 47 –

* MIR → Medium Infrared.

* NASA → National Aeronautics and Space Administration.

* ESA → European Space Agency.

* JAXA → Japan Aerospace Exploration Agency.

* COBE → Cosmic Background Explorer.

* HIPPARCOS → HIgh Precision PARallax COllecting Satellite.

* HST → Hubble Space Telescope.

* XMM → X-ray Multi-Mirror Mission Observatory - Newton.

* INTEGRAL → International Gamma Ray Astrophysics Laboratory.

* WMAP → Wilkinson Microwave Anisotropy Probe.

* FGST → Fermi Gamma-ray Space Telescope.

* GLAST → Gamma-ray Large Area Space Telescope.

* GAIA → Global Astrometric Interferometry for Astrophysics.

* VLA → Very Large Array.

* VLT → Very Large Telescope.

* VERITAS → Very Energetic Radiation Imaging Telescope Array System.

* SALT → South Africa Large Telescope.

* GTC → Gran Telescopio Canarias.

* ALMA → Atacama Large Millimeter Array.

* FAST → Five-hundred-meter Aperture Spherical radio Telescope.

* MAGIC → Major Atmospheric Gamma Imaging Cherenkov telescope.

* HESS → High Energy Stereoscopic System.

20 What are some of the astrophysical sources of radio waves ?

Quick Answer

Planets, Stars , Pulsar , Gas clouds, Comets, Galaxies, Cosmic background,

and Supernova remnants.

More Details

Depending on the source we distinguish two types of radio emissions: thermal

and non-thermal. Below I discuss with some details each one of them.

– Thermal emsission

– Non-thermal emission

– 48 –

21 What is the Ozone layer ?

Quick Answer

• Ozone is a molecule composed of three atoms of oxygen (O

)

• 10% in the troposphere, and 90% in the stratosphere

It was discovered in 1840 in Munich by the German chemist C. .F. Schonbein.

More Details

Figure 11: The layers of the Earth’s atmosphere. Source: Reference[17].

• Ozone formation and destruction

The formation of ozone molecules in the stratosphere (see Fig.. ??) is produced

in a two step process: In the ﬁrst step, UV photons from the Sun

strike the

molecules

in the stratosphere, dissociating some of them into individual

oxygen atoms; b) (Almost all) the produced oxygen atoms, O, combine with

oxygen molecules to produce O

molecules. The ozone generation also get

UV radiation constitute about of 5% of the Sun’s radiation.

which make up about 21% of the atmosphere.

– 49 –

destructed by the UV photons as well via its collision with oxygen atoms, which

keeps the concentration of the ozone in balance. This cycle of the formation

and destruction of the ozone molecules is known as Chapman mechanism

represented by the following chemical reactions:

Reaction 1 : O

+ γ

[λ

< 240 nm] → 2 O (21.1)

Reaction 2 : O + O

+ M → O

+ M (21.2)

Reaction 3 : O

+ γ

[λ

< 320 nm] → O

+ O (21.3)

Reaction 4 : O + O

→ 2O

(21.4)

Here M denotes a non-reactive particle that can take up some of the energy

released in the reaction

, γ

denotes a UV photon, and λ

is its wavelength.

So the net eﬀect of the above process is an equilibrium between atomic oxygen,

ozone, and diatomic oxygen, which can represented by the reaction

O + O

→ 2 O

(21.5)

As a result of this balance between the amount of ozone produced and de-

stroyed, its total concentration remains relatively constant at a given latitude

and altitude in the stratosphere. The greatest concentrations are at altitude

between 15 km and 25 km, forming ozone layer

and most of it is formed over

the equator where the amount of radiation from the Sun is higher.

• Impact of ozone depletion

Scientists commonly subdivide the ultraviolet light into three wavelength ranges

classiﬁed as UV-A , UV-B, and UV-C. The UV-C rays are the most dangerous

present the highest is completely absorbed by the ozone layer, whereas UV-

A and UV-B penetrate the atmosphere and reaches the Earth’s surface, with

about 95% is of the type UV-A, as can be seen in Fig. 12. Although it is

believed that UV-A rays they do not cause signiﬁcant skin damage over short

time exposure, they contribute to skin aging and wrinkles, and they increase

skin cancer risk when exposed to large amount. The UV-B can cause sunburn

and damage skin’s cell DNA. Thus, a decrease in the ozone layer would have

serious eﬀect on life on Earth, in particular due the increase of the intensity of

UV-B radiation at the Earth’s surface.

In honor of Sydney Chapman who proposed it in 1930.

M is usually N

or O

The ozone layer was discovered in 1913 by the French Physicists C. Fabry and H. Buisson.

Although these divisions has its grounding in Physics but they turn out to be very useful in

biology.

– 50 –

Figure 12: The three types of UV radiations and ozone layer.

• Causes of ozone depletion

One of the major causes of ozone depletion in the stratosphere is chloroﬂuo-

rocarbons, CCl

, usually abbreviated as CFCs, which is used in many ap-

plications, including refrigeration and air conditioning. The reason for the

CFC being a source of ozone-depletion is the presence of chlorine

atoms that

catalyzes the breakdown of ozone to diatomic oxygen via the reaction

CCl

+ γ

→ C ClF

+ +Cl

CCl

F + γ

→ C Cl

F + +Cl

+ Cl → ClO + O

ClO + O → Cl + O

A CFC molecule does not interact with anything in the lower atmosphere and

eventually reaches the upper atmosphere where the Sun’s radiation is intense

enough to break it apart producing chlorine. Moreover, since the CFC has a

lifetime of about 20 to 100 years, from the chain reaction above, at very low

temperature, as it is the case over the Antarctic, a single chlorine atom can

Actually, the radicals nitrogen oxide (NO), hydrogen oxide (HO), and bromine oxide (BrO),

are also eﬀective ozone-killers.

– 51 –

destroy hundred thousands of ozone molecules before it form one of other chlo-

rine compounds and exit the stratosphere. In fact, in 1985, British scientists

discovered that the ozone layer over the Antarctic was thinning. This ozone

decline over the south pole was called ozone hole

Fortunately, in 1987 many industrial nations came together and agreed to phase

out the CFCs and signed the Montreal protocol, and over the next decades most

of the world’s nations have signed up.

22 What is a parallax angle? a parsec? a light year ?

Quick Answer

• In astronomy, the parallax is half the angular displacement of nearby star

as seen from the Earth at one time of year, and six months later.

• A parsec, denoted by pc, is the distance to an object that has a parallax

angle of 1 arsecond.

• A light year is the distance that light travels in one year.

• The relation between parsec and light year is given by

1pc = 3.26 ly

More Details

• Parsec (pc), astronomical unit (AU), and light years (ly)

If D denotes the distance to an astronomical object, and p the the parallax

angle, then we have

D =

average distance between the Earth and the Sun

1 AU

p[in radian]

= 1 AU



arsec



arsec



(22.1)

60 × 60 × 180

× 1 AU



arcsec



(22.2)

Thus,

The ﬁrst ozone hole over the Arctic was observed in 2011 after an unusually cold Arctic winter.

– 52 –

D = 206, 265 AU



arcsec



(22.3)

Now, since by deﬁnition 1 pc is the distance that corresponds to a parallax

angle p = 1 arcsec, we deduce

1pc = 206, 265 AU (22.4)

To express distances in light years, we use the fact that

Figure 13: Parallax.

1 ly = speed of light × one year time interval

= 2.99 × 10

m/s × 3.15 × 10

' 9.42 × 10

Hence, we obtain

D =

206, 265 AU

9.42 × 10



arsec



ly = 3.26 ly



arsec



Setting p = 1arcsec in the above expression results in a distance of 1 pc, from

which it follows that

1pc = 3.265 ly ' 3 × 10

m (22.5)

• The ﬁrst Measurement of parallax of a star

In 1838, the German astronomer Friedrich Bessel, successfully measured the

parallax of a star, and that was for the star 61 Cygni in the constellation of

Cygnus. He obtained p

(61Cygni)

= 0.286 arcsec, which corresponds to a distance

of 11.4 light years from Earth.

– 53 –

• Measuring the parallax of Mars

In 1972, Cassini and with his colleague Jean Richer, made a simultaneous

measurement of the position of Mars at opposition

from Paris (France) and

Cayenne (French Guiana), respectively, and determined that the parallax angle

of Mars

with respect to Paris-Cayenne baseline to be

p[Mars] ' 25 arcsec =

25 × π

180 × 3600

= 1.2 ×10

−4

radian

Given that the distance between Paris and Cayenne is 6437 km, he deduced

that the distance between Earth and Mars is

Earth→Mars

distance between Paris and Cyenne

p[Mars]

= 6437 ×

180 × 3600

24 × 2π

' 55 million km (22.6)

The measurement of the Earth-Mars distance can be used to calibrate the

astronomical unit AU in kilometers, and hence determine the distances from

the planets to the Sun. Cassini Knew that at opposition Mars is 3/8 AU from

Earth, and hence he deuced that

Earth→Sun

' 140 million km (22.7)

which is only 7% smaller than the accepted modern value.

• Measuring distances to other stars

– Earth Based Telescopes

: By using adaptive optics to remove the

distorting eﬀect due to the atmosphere

, it is possible to measure paral-

lax angles as small as 0.01 arcsec, corresponding to a distance of 100 pc.

– Space Based Telescopes: Being in space, these telescopes avoid many

of the observational problems associated with ground-based measure-

ments, and can measure distances to objects up to about few ×100 pc

away. For instance, the Hipparcos satellite

was able to measure the

parallax of more than 100, 000 stars with a precision of 0.001 arcsecond,

It means when it was on the opposite side of the Earth from the Sun, and hence closest to

Earth.

Venus can be in inferior conjunction with Sun, i.e. on a line between the Earth and the Sun,

and as a result it can be closer to the Earth than Mars. So one would expect its parallax to be

larger than that for Mars, but unfortunately it is usually invisible in such position.

With ordinary telescope, one can observe stars at distances about 30 pc ' 100 ly away from

us.

This eﬀect is known among astronomers as the "seeing" eﬀect.

It was launched by the European Space Agency (ESA) in 1989 and operated until 1993.

– 54 –

Name Parallax [arcsec] Distance [lyr] Distance [pc]

Proxima Centauri 0.770 4.24 1.30

Alpha Centauri A & B (binary) 0.750 4.37 1.33

Bernard’s star 0.546 5.96 1.83

Wolf 359 0.419 7.86 2.39

Lalande 21185 (binary) 0.395 8.31 2.53

Sirius A & B (binary) 0.382 8.66 2.62

Luyten 726 − 8 0.374 8.73 2.68

Ross 154 0.345 9.68 2.90

Ross 248 0.316 10.32 3.16

Epsilon Eridani 0.305 10.70 3.28

Table 6: The distance of the 10 closest stars to our solar system

corresponding to a distance of a kpc, which is about 1/30 the size of our

galaxy (see Q9).This is in addition to about million stars that Hippar-

cos measured their distances, but with lower accuracy. The successor of

Hipparcos, the Gaia satellite (see Q12), measured parallaxes with an

accuracy of 4 µarcsecond for about a billion stars down to 20th magni-

tude ((see Q19)), giving distances up to 25 kpc with 10% accuracy

In table 6 we show the parallax angles and the corresponding distances

to the ten closest stars to Earth.

23 What is the mass of the Moon ?

Quick Answer

7.4 ×10

kg ' 1% M

⊕

More Details

• Newton’s estimate

The ﬁrst attempt to estimate the mass of the Moon was by Isaac Newton by

using the measurements of the sea tides. He showed that the height the tides

produced by an astronomical body is proportional to its mass, and determined

that the ratio of the mass of the moon to the mass of the earth is approximately

The ﬁnal Gaia catalogue is expected to be released around 2022.

– 55 –

1/40, which is 50% smaller than the actual value. The details of this ﬁnding

are a bit complicated but we will discuss with some details the tidal forces and

their eﬀects on the Earth in the next section.

• An estimate

Let us consider the Moon to have a spherical shape with radius R

, and density

, then its mass is given by

Moon

4π

(23.1)

Since it is widely accepted that the Moon formed out the debris left over from

the impact of Mars-sized astronomical object with the Earth approximately

4.5 billion years ago

the Moon seems to be formed out Earth, it is reasonable

to assume that ρ

is similar to the density of a typical Earth’s rocks, i.e.

∼ 3 g/cm

and use the fact that R

is about 27% the radius of the

Earth

, we obtain

Moon

∼ 7 × 10

kg (23.2)

• Better estimate

24 What is stellar magnitude?

Quick Answer

This is known as the Giant-impact hypothesis. The Mars-sized that collided with the

proto-Earth was named Theia, which Greek mythology is the mother of Selene, the

goddess of the Moon.

The estimate of the size of the Moon was known since the ancient Greek (more than 2000 years

ago) from the timing the lunar eclipse. See also A 2.

– 56 –

We deﬁne L be the intensity of the electromagnetic radia-

tion emitted by an astronomical object

, i.e. the amount of

energy the object radiates per unit time. For an observer,

the energy detected per unit time and unit area, denoted by

F, is the ﬂux of the observed radiation.

• Apparent magnitude of star X [m]:

(X)

= −2.5log



(X)

(Vega)



• Absolute magnitude of star X [M]:

(X)

− M

(X)

= −2.5 log



10 pc



Here d is the distance from the star X to the observer.

It is also known as the luminosity the absolute brightness.

More Details

• Apparent magnitude and absolute magnitude

Around 150 BC, the Greek astronomer Hipparchus classiﬁed stars by their

brightness, with the brightest were termed ﬁrst magnitude, next brightest

were second magnitude through to the sixth magnitude which corresponds

to the faintest stars a naked eye can see

. Hipparchus magnitude scale-system

reﬂected the fact that when the intensity of a light source increases linearly,

human eye’s response is not linear, instead the eye sensitivity is logarithmic.

• With the invention of the telescope, astronomers were able to see much fainter

stars which lead them to extend the magnitude scale system and decided to

formulate it in a more precise way. For that, in 1856, the English astronomer

Sir Norman R. Pogson proposed to deﬁne a 5 magnitude diﬀerence between

two stars corresponds to a ratio of 100 to 1 in brightness. In general, if two

stars A and B have apparent magnitudes m

(A)

. and m

(B)

, respectively, and

(A)

and F

(B)

are corresponding ﬂuxes, then

(A)

(B)

= 10

0.4

(

(B)

−m

(A)

)

⇐⇒ m

(A)

− m

(B)

= −2.5 log



(A)

(B)



(24.1)

This ranking is counter intuitive as one expect the magnitude to increase with the increase of

brightness.

– 57 –

The magnitude system was extended to negative values to accommodate very

bright objects

. By convention, astronomers use the star Vega

as a refer-

ence and take its apparent magnitude equal to zero

. Thus, the apparent

magnitude of a star X can be expressed as

(X)

= −2.5 log



(X)

(Vega)



(24.2)

This process of measuring the apparent brightness of an object. is called pho-

tometry

Note that the apparent magnitude of an object depends on its dis-

tance to the observer, and therefore, in order to be able to compare their

luminosity, astronomers introduced the absolute magnitude scale, denoted

by M, which is deﬁned as the brightness that a star appears to have if it were

placed 10 pc = 3 × 10

m (see A 17) away from the observer, i.e

m − M = −2.5 log



(observed)

(at d = 10 pc)



(24.3)

So the absolute magnitude indicates the star’s intrinsic brightness, i.e. it is

independent of how far the star is from the observer. Now, we recall that for a

source of known luminosity L, which emits radiation isotropically at a distance

d from an observer, the ﬂux F received reads

F =

4πd

(24.4)

which we substitute into (24.3) and get

m − M = 5 log





− 5 (24.5)

The quantity (m −M) is called the distance modulus and represents a mea-

sure of distance to the object.

• Filters

∗ Because stars have diﬀerent temperatures, and hence they emit radiation

strongly in diﬀerent wavelength ranges, instruments that measure the

The magnitude system is useful only for wavelengths in the visible, UV, and infrared spectrum.

For other wavelengths astronomers use the ﬂux units: Jansky (1Jy = 10

−26

W/m

/Hz for radio

waves, and erg/cm

/s for X-ray, and gamma-ray.

The star vega is at a distance of about 25 lys from Earth, and it is part of the constellation

Lyra in the northern hemi-sphere.

Because in the original magnitude classiﬁcation the star Vega was one of the brightest star.

The maximum apparent magnitude of a naked human eye is +6 in the best observing conditions,

and +3 in an urban environment. In contrast, The Hubble Space Telescope (HST) can reach

apparent magnitude of +30 in the visible band

– 58 –

ﬂuxes arriving from a star have ﬁlters that can allow only some narrow

range (or band) of wavelengths to pass through. The amount of spec-

trum that a ﬁlter allows to pass is called the bandpass and the central

wavelength of the band pass is known as the eﬀective wavelength. We

denote by m

(or sometimes by just a capital letter K) and M

the

apparent and absolute magnitudes in the spectral K-band, respectively.

∗ The most widely used ﬁlter systems are the [U, B, V, R, I] in the optical

and the [Y, J, H, K] in the near-infrared. For instance, the V ﬁlter allows

only light in the wavelength range [505 −595 nm] to pass through the

ﬁlter

∗ In deﬁning the magnitude at some wavelength λ one needs to choose the

zero-point of magnitude. For a long time astronomers commonly used the

Vega- magnitude system, in which m

(Vega)

= 0 in all passband

So, the apparent monochromatic magnitude of an astronomical object

reads

Vega;λ

= −2.5 log

(Vega)

(24.6)

where F

is the ﬂux per unit wavelength. Thus, the zero point of this

system depends on the ﬂux of the star Vega and is diﬀerent in diﬀerent

bands. For example, in the V passband, we have

(Vega; V-band)

= 3.63 ×10

−9

erg.s

−1

.cm

−2

−1

(24.7)

In terms of frequency, we have

|dν| = F

|dλ| =⇒ F

where in the second equality we made use of the relation λ = c/ν, where c

is the speed of light. Thus, for the star Vega, the zero pointy ﬂux density

in the V-band is

(Vega; V-band)

= 3.63 ×10

−20

erg.s

−1

.cm

−2

.Hz

−1

In table 7. we give the eﬀective wavelength (λ

eﬀ

), the bandpass (∆λ), and

the averaged ﬂux density (F

) in each band of the UBVRI photometric

system.

This ﬁlter system was introduced in 1953 by H. Johnson and W. Morgan (see ref. [21]).

In day light human eye is most sensitive to radiation with wavelength of about 550 nm, and it

decreases towards the red.

The Vega-magnitude system was generalised to a class of stars called A0V stars.

– 59 –

Filter λ

eﬀ

[nm] ∆λ [nm] F

[W/m

/Hz

−1

]

U 300 50 1.81 ×10

−23

B 430 72 4.26 × 10

−23

V 550 86 3.64 ×10

−23

R 650 133 3.08 × 10

−23

I 820 140 2.55 × 10

−23

Table 7: The characteristics of the UBVRI ﬁlters in the Vega-magnitude system.

∗ Another magnitude system which is becoming increasingly popular is the

AB-magnitude system deﬁned as

AB;ν

= −2.5 log



(3.6308 × 10

−23

W/Hz/m

)



(24.8)

or, equivalently,

AB;ν

= −2.5 log

− 48.6 (24.9)

where F

is in erg.s

−1

.cm

−2

.Hz

−1

unit. Note that in contrast to the Vega-

magnitude system no stellar spectral distribution is used as a reference,

and instead the zero-point ﬂux is a constant ﬂux at all frequencies, and

was chosen in such away AB-magnitude and Vega-magnitude are the

same at the wavelength λ = 5500

∗ An example: The apparent magnitude of the Sun in the V band reads

()

= −2.5 log

()

(Vega)

(24.10)

Using the value of F

(Vega)

in (24.7) and the measured solar ﬂux in the

V-band F

()

= 185 erg.s

−1

.cm

−2

−1

, we get

()

= −26.7 (24.11)

Now substituting the above value of m

()

and the mean Earth-Sun dis-

tance (d = 1 AU ' 4.85 × 10

−6

pc) into Eq (24.5), we ﬁnd

()

= +4.83 (24.12)

Sometime it is useful to compare the magnitudes of stellar objects with

that of the Sun, which can be expressed as

100

− M

()

= −2.5 log



K,



(24.13)

Note that the AB-magnitude system is deﬁned in terms of the ﬂux per unit frequency.

100

The absolute magnitudes of the Sun in diﬀerent ﬁlters can be found in [? ].

– 60 –

Celestial Object distance [pc] m

L/L



Sun 4.85 × 10

−6

−26.7 +4.83 1

Moon 1.25 ×10

−8

−12.6 It is not a star It is not a star

Venus 3.51 × 10

−6

−4.6 It is not a star It is not a star

Proxima Centauri 1.29 +11.0 +15.5 5.4 × 10

−5

Alpha Centauri A 1.32 −0.01 +4.4 1.52

Alpha Centauri B 1.32 +1.33 +5.7 0.5

Sirius A 2.64 −1.6 +1.45 23.6

Polaris 133 +2 −3.6 10

Betelgeuse 197 +0.41 −5.2 10

Rigel 265 +0.14 −6.8 4.5 × 10

Table 8: The brightness of some celestial objects and their distance from Earth.

In table 8 we list some some celestial objects with their corresponding

apparent magnitude, absolute magnitude, and their luminosity relative

to the Sun’s luminosity.

∗ If the ﬂux of radiation is measured across all range of the electromagnetic

spectrum, then the magnitude is called the bolometric magnitude,

and denoted by m

bol

and M

bol

for the apparent and absolute magnitude,

respectively.

∗ Color indices

The diﬀerence in magnitude between two ﬁlters, which is equivalent to

measuring the ﬂux ratio passing through the two ﬁlters, is called a color

index. The color index of a star is an indication of its surface tempera-

ture

101

→ hot stars have (B − R) < 0, they are bright in the blue, and faint in

the red,

→ cool stars have (B − R) > 0, i.e. they are faint in the blue and bright

in the red.

By approximating stars as black bodies, Ballesteros (see ref. [23]) de-

rived a formula of a star’s surface temperature in terms of the color index

(B − V ), given by

(surface)

= 4600 K



0.92(B − V ) + 1.7

0.92(B − V ) + 0.62



(24.14)

101

Another way to estimate the surface temperature of a star is to use Wien’s displacement law

(see A 23).

– 61 –

For instance, applying the above formula for the Vega star, which by

deﬁnition has color index zero, yields T

(surface)

Vega

∼ 10, 000 K.

• The Eﬀect of Atmospheric Absorption

The deﬁnitions of the magnitudes discussed previously assumed that the ﬂuxes

are measured in the absence of the Earth’s atmosphere or at the top of it. So for

observations performed on the Earth, the incoming electromagnetic radiation

suﬀer. absorption and scattering, which result in atmospheric extinction of the

ﬂuxes at the Earth’s surface. Furthermore, the interstellar space is not perfectly

empty as it contains gas and dust grains, and hence there will be attenuation of

the observed electromagnetic ﬂux radiated by the stellar objects. These eﬀects

can be corrected by redeﬁning the distance modulus as

102

= M

+ 5 log



10 pc



+ A

(24.15)

where the extinction term A

= 1.086τ(λ), with τ(λ) is called the optical

depth and is given by

n(r

)σ(r

, λ) dr

(24.16)

Here, n(r) is the number density of the particles in the medium at distance r

from the observer and σ(r, λ) is the scattering cross section

103

of a photon of

wavelength λ on a particle of the medium.

102

Eq (24.15) is the consequence of the deﬁning equation of the extinction term, i.e.

= m

λ;0

+ .A

with m

λ;0

is the apparent magnitude in the absence of any medium of particles between the stellar

object and the observer.

103

The cross section is proportional to the probability for the scattering to occur.

– 62 –

25 What is the average solar power density at the Earth’s atmosphere?

Quick Answer

1368 W/m

More Details

The value of the solar power density reaching the surface of the Earth’s atmosphere

is known as the solar constant, and it is usually denoted by G

. If there was no

atmosphere, the total power of solar radiation reaching the Earth?s surface would

be the solar constant times the cross sectional area of the Earth

104

, which yields

1.7 × 10

Watt. Thus, in one year, the top of the Earth?s atmosphere receives on

average approximately 5.46 Yotta Joules = 1.52 × 10

kWh

105

. Hence, the mean

irradiance at the surface of the Earth is

(0)

Fraction of sunlight incident of the Earth

Area of the Earth facing the Sun

1.7 × 10



2 × 3.14 × (6, 400 × 10



' 680 W/m

(25.1)

Because Earth has an atmosphere, around 30% of the solar energy gets reﬂected

by the clouds into space and 20% is absorbed by water vapor, dust and O3

molecules in the air. Thus, the solar energy incident on the Earth

106

per day is

107

with atm

atm

= 50% (25.2)

This means that even if just 0.1% of the solar energy that reaches the ground is used

it can power the entire world.

26 How much electromagnetic power is radiated by the Sun?

Quick Answer

• The amount electromagnetic energy radiated per second by an object

X is called the luminosity of the object, and it is usually denoted by L

• For the Sun, L



= 3.9 × 10

104

That is πR

⊕

, with R

⊕

∼ 6, 400 km is the radius of Earth.

105

Units conversions: Yotta = 10

; kWh = 3.6 × 10

Joule.

106

Also called the insolation.

107

Assuming that the day is an average 12 hours at any location on the Earth.

– 63 –

More Details

Since we know the Earth-Sun distance, d

E-S

' 1.5 ×10

m (see A

3) , and the solar

constant, G

(0)

= 1368 W/m

(see A

21), the Sun’s luminosity is given by



= G

(0)

× 4πd

E-S

' 3.9 × 10

W (26.1)

27 How much is the surface temperature of the Sun?

Quick Answer

(surface)



' 5800 K

More Details

• The observed electromagnetic radiation emitted by the Sun has approximately

a blackbody spectrum, and hence we can apply Wien’s displacement law

108

max

T = 0.29 cm.K (27.1)

where λ

max

is the wavelength in the spectrum with the maximum intensity. For

instance, the sun emits maximum radiation at the visible yellow wavelength,

which corresponds to λ ∼ 500 nm, and hence we ﬁnd

T ' 5800 K (27.2)

• This method of estimating the surface temperature can be also applied to stars

as long as the spectrum of emitted the radiation is a Planck function. However,

in reality the electromagnetic spectrum from stars contains absorption lines,

and so it may not be possible to observe a peak in the spectrum. Moreover,

the interstellar dust may distort the spectrum light, an eﬀect known as the

interstellar reddening, which can produce erroneous temperature

109

. Instead,

astronomers use absorption lines to determine spectral classiﬁcation.

108

One can measure the light spectrum emitted by the star, ﬁt it with Planck?s distribution and

deduce the temperature.

109

Applying Wien’s law to a reddened spectrum will yield a temperature that is too cool.

– 64 –

28 Is there water on Mars?

Quick Answer

• Yes, but in icy form on its surface at the north polar ice cape and

small quantities as vapor in its atmosphere.

• With the current atmospheric pressure at the surface of Mars (6 −

14 millibar)

, most the ice sublimes

(see below for details).

• There are some recent evidence

for liquid water on Mars deep underground.

1 bar = 0.987 atm.

It means ice turns directly into water vapor without ﬁrst transitioning into a liquid.

2019

More Details

• Water can exist in three states: solid, liquid, and gas. At certain temperature

and pressure, water can exists in the three phases simultaneously. This is

achieved at temperature of 273.16 K and pressure 0.006 atm. This point is

called the the triple point of water and is shown in Fig. 15 for the pressure-

temperature phase diagram

110

• From Fig. 15 we see that the atmospheric surface pressure on Mars is very close

to the triple point of water, and this implies that is it is hard for liquid water

to exist. Thus, in the present-day Martian surface liquid water is not stable

and can be only be in a solid form (i.e. ice) or in gas form (i.e. vapor).

• However, at lower elevation. such as area created by impact of a large asteroid

the pressure can be higher than the the pressure of the triple point of water and

in this case if the temperature is about right (according to the phase diagram),

then transition from ice to liquid water may be possible if the term. For

instance, the pressure at the Hellas Basin is 12 mbar and hence in principle

liquid water could form.

• In 2018, Mars Express orbiter [2003- present], reported the detection of liquid

water, 1.5 km below the southern polar caps and expanding 20 km horizontally.

110

Note that the curve separating the solid and liquid phases is tilted to the left (in reality it is

slightly tilted). This implies that ice is less dense than liquid water; this is why icebergs ﬂoat on

the oceans surface, and also this is why lakes and rivers freeze from top to bottom, allowing the

life-form to survive even when the surface has frozen.

– 65 –

Figure 14: Schematic phase diagram for water. The atmospheric surface pressure

on Earth and on Mars are shown in blue and orange colors, respectively.

Name Launching Date Agency Type Operational Until

Mars Odyssey 2001 NASA Orbiter 2025

Mars Express 2003 ESA Orbiter 2026

Opportunity 2003 NASA Rover 2018

Mars Reconnaissance 2005 NASA Orbiter 2030s

Curiosity 2011 NASA Rover + 2019

ExoMars Trace Gas Orbiter 2016 ESA/Roscosmos Orbiter + 2020s

Table 9: Some of successful mission to Mars.

The discovery has been made using Marsis

111

, a radar instrument placed on

board of the Mars Express satellite. Marsis bounces oﬀ radio waves oﬀ a

selected area of the planet and then analyzes the received echoes.

• Successful missions to Mars in the last 20 years[See table 9]

111

It stands for Advanced Radar for Subsurface and Ionosphere Sounding instrument which was

developed and built by Italian scientists from the the European Space Agency (ESA).

– 66 –

29 How do we know if a star or a galaxy is moving toward or away from us?

Using the Doppler shift ......

More Details

• Doppler eﬀect for sound waves

112

Let us denote T

emit

the period of wave at the source, λ

emit

the wavelength of the

wave at the source, λ

rec

the wavelength of the wave detected by the observer,

the speed of the sound wave, υ

src

the speed of the source, υ

rec

the speed of

the observer, d the distance between the observer and the source at the instant

when the ﬁrst crest is emitted by the source (i.e. at t = 0), t

rec

is the instant

at which the ﬁrst crest is received by the observer, t

rec

is the instant at which

the second crest is received by the observer.

– Source moving and the observer standing still

– Observer moving and the source standing still

• Doppler eﬀect for EM waves

30 What is the spectral classiﬁcation of stars ?

Quick Answer

• The strengths of the hydrogen lines depends on the surface temperature

of the star.

• Hence, the apparent color of a star gives an approximate estimate of its

surface temperature.

• The order of spectral types of stars from hottest to coolest is labeled

OBAFGKM

To remember this order one can use the following mnemonic: Oh Be A Fine Girl Kiss Me

More Details

112

In 1842, the Austrian Mathematical Physicist Christian Doppler wrote an article on the

behavior of sound waves and showed that the relative motion of the source and the observer, the

detected and the emitted frequencies of the wave are diﬀerent.

– 67 –

31 What is the Eddington limit ?

∼ XXXX

More Details

•

32 What is the minimum temperature for hydrogen fusion inside a star ?

∼ 10 million degrees

More Details

• The estimate of temperature from classical Physics

The nuclear fusion of hydrogen is the process in which protons combine to form

helium nucleus. This can occur via the strong nuclear force, which has has a

range of about the size of a medium nucleus, i.e.

strong

∼ 10

−13

cm (32.1)

Since electric charge of nuclei is of the same sign (positive), there will be an

electrostatic potential energy barrier

Coulomb

strong

' 2 × 10

−13

J = 1.2 (Z

) MeV (32.2)

To overcome this Coulomb barrier, the protons must have large kinetic energy,

which inside a star it corresponds to a very high temperature. So if we consider

a gas of protons in the core of a star. at temperature T , then according to

kinetic theory of gases the average kinetic energy per particle is proportional

to the absolute temperature, and it is given by

< K

> =

T ' keV





(32.3)

However, as we will show in A 26, the temperature in the core of a typical

star, such as the Sun, is of order 10

K (for details see A XX), and hence the

protons can not fuse in such a star nuclear. In fact, even for T ∼ 10

K, the

particles (protons) will not have energy to penetrate of the Coulomb barrier,

and hence the nuclear fusion can not occur; at least classically.

– 68 –

• Estimate of the temperature using quantum mechanics

The solution to the above problem is quantum physics, where microscopic

particle can have wave-like properties; known as wave- particle duality, which

was suggested in 1921 by Louis de Broglie. It states that matter behave both

like particles as well as waves. For a particles with momentum p we associate

a wave with the wavelength

(32.4)

Thus, for the nuclear fusion between two nuclei to occur they only need to be

within distances that are comparable to their de Broglie wavelengths. Now

we require the kinetic energy of the two interacting particles be equal to the

Coulomb potential at a separation distance of one de Broglie wavelength,

µυ

2µ

2µλ

(32.5)

from which we infer the distance the particle need to be of one another

2µZ

(32.6)

Substituting the λ

into the expression of the Coulomb potential yields

Coulomb

(32.7)

Now, to determine the minimum temperature needed for a nuclei to penetrate

the Coulomb barrier we set the above relation of E

Coulomb

equal to the thermal

average kinetic energy per particle given in (32.3). We get

min

= 0.96 ×10







(32.8)

which is of order the central temperature of a typical star like our Sun. Hence,

we expect proton-proton fusion to take place in the core of the Sun

113

The above estimate for the fusion temperature seems to suggest that at T ' 10

million Kelvin any pair of protons will fuse. However, as mentioned above, the

process of fusion is aided by the quantum mechanical eﬀect mentioned above,

and hence the correct way to estimate the interaction rate of such process which

is proportional to the probability for a nuclei to go through the Coulomb barrier.

According to quantum mechanics, the charged particle has a non-zero trans-

mission probability to go through the electrostatic barrier even if the thermal

113

As we will see in A XX the fusion of hydrogen through a series of steps.

– 69 –

kinetic energy is order of magnitudes smaller than what is required to overcome

it. This phenomenon is known as the quantum tunneling. The derivation of the

tunneling probability and the interaction rate will be discussed in Appendix X.

33 What is the smallest mass that a star can have?

• M

∗

≥ 0.08 M



(i.e. at least 80 times Jupiter mass).

• Below this minimum value there is not enough thermal energy to start

the nuclear fusion of hydrogen in the star’s core

• The "failed stars" (i.e. with M < 0.08 M



) are called "brown dwarfs".

Some astrophysicists consider the 13 M

Jupiter

is the appropriate cut oﬀ between a star and

planet as a brown dwarf with mass larger than this cut oﬀ can burn deuterium (a.k.a heavy

hydrogen; with nucleus having one proton and one neutron) producing some small amount of

energy.

More Details

• Electron degeneracy.

• Estimate of minimum mass of a star

34 What is the largest mass a star can have?

• M

∗

≤ 120 M



• Above this maximum value there will be excessive energy production that

prevents any further mass accretion. In this case hydrostatic equilibrium

can not be maintained.

• .

More Details

– 70 –

35 What is a white dwarf?

• It is the stellar core remnant of star with mass M

∗

< 8 M



• It is mostly helium. It can also contain carbon and oxygen.

More Details

36 What is a black hole (BH)?

A region of space from which nothing, including light, can escape, which is

why we perceive them as black.

More Details

• Historical background [18

− 20

century]

– It was the English philosopher, geologist and Clergyman John Michel who

in 1784 proposed the concept of dark star, an object compressed into

suﬃciently small size that it would have an escape velocity that exceeds

the speed of light.

– The size that an object of mass M would have for it to be a BH is

known as the Schwarzschild radius can be determined using Newtonian

mechanics by setting the object’s escape velocity υ

esc

= 2GM/r equal to

the speed of light, c, and we obtain

2GM

' 3 km







(36.1)

For instance, to make a black hole out of the the Sun it has be compressed

to a radius of 3 km, and an Earth mass black hole has to have a radius

of approximately 1 cm. However neither the Earth nor the Sun have the

necessary gravitational force to overcome the electron and the neutron

degeneracy pressures, respectively.

– According to the theory of general relativity (GR), introduced by Albert

Einstein in 1915, a BH is a result of the curvature of spacetime caused

by a huge mass.

– 71 –

– In 1915, Karl Schwarzschild provided the ﬁrst exact solution to the Ein-

stein ﬁeld equations of GR, for the case of a spherical non-rotating mass.

He showed that the points at the Schwarzschild radius surface, i.e. r = r

deﬁnes the the event horizon beyond which light can not escape.

– The term "black hole" was coined by the American physicist John Wheeler

in a lecture he gave in 1967.

– The ﬁrst BH to be discovered was in 1971, named Cygnus-X1, located

in the Cygnus constellation at about 6000 lyrs from us. It turns out

that Cygnus X-1 is a source of X-ray

114

and form a binary system that

includes blue supergiant variable star

115

which it orbits at about 0.2 AU.

– In 1974, two astronomers, Bruce Balick and Robert L. Brown, published

a paper describing a bright radio source in the constellation Sagitarius

at the very center of the Milky Way. It turns out this radio source is a

supermassive black hole in the center our galaxy, and was named Sgr

A* black hole (see Fig. ??). Observation of other galaxies showed they

too have supermassive black home at their center.

• Schwarzschild solution

• The Tolman-Oppenheimer-Volkoﬀ equation

• Near the black hole

37 How are black hole (BH) discovered?

By studying the way matter moves in its vicinity. For example, X-ray emission

.....

More Details

114

This was actually ﬁrst spotted in 1964 during a rocket ﬂight.

115

It was designated HDE 226868.

– 72 –

Figure 15: The inferred orbits of Stars (the red and yellow ellipses show the orbits

of SO-2 and SO-102 stars, respectively) orbiting the supermassive black hole in the

center of our galaxy [Credit: Andrea Ghez and colleagues using the Keck Telescopes].

38 What is the Event Horizon Telescope (EHT)?

• The EHT is a large telescope array, consisting of a global network of radio

telescopes, forming an Earth-sized virtual telescope.

• The ﬁrst image of the event horizon of a black hole.

More Details

The EHT collaboration involves more than 200 researchers from Africa, Asia, Europe,

and North and South America, from the following radio observatories:

– 73 –

Telescope Country Wavelengths Comments

ALMA Chile 0.3 −3.6 mm Interferometer of 66 radio telescopes

(54 antennas with 12 m diameter)

APEX Chile 0.2 −1.5 mm A single dish of 12 m diameter

SMT USA sub-mm A single dish of 10 m diameter

JCMT USA 0.4 −1.4 mm A single dish of 15 m diameter

IRAM 30m Spain 1 −3.7 mm A single dish of 30 m diameter

IRAM NOEMA France sub - few mm Interferometer of 8 radio telescopes

(each antenna has 15 m diameter)

The acronyms:

* ALMA → Atacama Large Millimeter Array.

* APEX → Atacama Path Finder Experiment.

* SMT → Submillimeter Telescope.

* JCMT → James Clerk Maxwell Telescope.

* IRAM → Institute for Radio Astronomy in the Millimeter Range.

* NOEMA → NOrthernExtendedMillimeterArray.

39 What is the resolving power of a telescope?

It is the ability of a telescope to show the ﬁne details of an object and diﬀer-

entiate the images of two objects close to each other.

More Details

40 What are the major components of a radio Telescope?

A radio telescope consists of three main components:

1. A radio reﬂector dish

2. A radio Antenna (receiver)

3. A recorder

More Details

– 74 –

41 What is the epoch of reionisation?

• It is the period when the universe changed from being mostly neutral to

mostly ionized.

• It is approximately 600 million years after the big bang.

• It is caused by the ﬁrst stars which are believed to have been very massive

and highly luminous at wavelength below 912 A

More Details

42 How many galaxies are there in our visible universe?

About two trillion galaxies

More Details

• Before the 20 th century it was thought that the universe is the Milky way,

until 1925 when Hubble .....

43 What is the closest known galaxy to the Milky Way?

Andromeda galaxy

More Details

•

– 75 –

44 What is the farthest known galaxy in the universe?

• The galaxy GN-z 11, discovered by the Hubble space telescope in 2018.

• It is at about 32 billion light years from us.

More Details

The light emitted by the GN-z 11 galaxy and detected by the Hubble space

telescope left this galaxy 13.4 billion years ago. Now one might wonder how come

.....

45 What is a neutron star?

XXXXXXXXXXXXXXXXXXXXXXX

More Details

46 How old is the universe?

XXXXXXXXXXXXXXXXXXXXXXX

More Details

47 How big is the universe?

XXXXXXXXXXXXXXXXXXXXXXX

More Details

48 What is the Cosmic Microwave Background?

XXXXXXXXXXXXXXXXXXXXXXX

More Details

– 76 –

49 What is the epoch of dark ages?

XXXXXXXXXXXXXXXXXXXXXXX

More Details

50 What is era of big bang nucleosynthesis?

XXXXXXXXXXXXXXXXXXXXXXX

More Details

– 77 –

51 Appendices

51.1 Tunneling Through Coulomb Barrier

• Rectangular Potential Barrier

We will ﬁrst consider the behavior of quantum particle at a one-dimensional

rectangular potential barrier of height V

(see Fig.)

V (x) =







0. x < 0

, 0 < x < a

0, x > a

(51.1)

We assume that the particle has energy E < V

, moving from the left toward the

potential. Thus, the solutions to the time-independent Schrodinger equation

in the regions x < 0 (region I), 0 < x < a (region II), and x > a (region III),

are given by

116

(x) = Ae

ikx

+ Be

−ikx

, (51.2)

(x) = Ce

−κx

+ De

κx

, (51.3)

III

(x) = Ee

−κx

(51.4)

Here k =

√

2mE/~ is the momentum of a free particle, κ =

2m (V

− E)/~,

the A-term is the incident wave moving to the right and the B-term is the

reﬂected wave moving to the left. Using the fact that wave function and its

derivative must be continuous at x = 0 and x = a, i.e. ψ

(0) = ψ

(0), ψ

(0) =

(0), ψ

(a) = ψ

III

(a), and ψ

(a) = ψ

III

(A) and ψ

(a) = ψ

III

(a), we obtain











A + B = C + D

ik (A − B) = −κ (C − D)

−κa

+ De

κa

= Ee

ika

−κ (Ce

−κa

− De

κa

) = ikEe

ika

(51.5)

Note that there four equations for ﬁve unknowns, and hence we can only ex-

press four the ﬁve variables in terms one of them. Our aim is to determine the

transmission probability of the particle through the potential barrier, which is

116

In one dimension, the time-independent Schrodinger equation reads

+ [E − V (x)] ψ = 0

– 78 –

deﬁned as the ratio between the magnitudes of the transition current probabil-

ity and the incident current probability

117

, i.e.

T (E) =



∗

III

(x)ψ

III

(x) − ψ

III

(x)ψ

∗

III

(x)

∗

(x)ψ

(x)

− ψ

(x)ψ

∗

(x)



|E|

|A|

(51.6)

By solving the system of equations (51.5) for the coeﬃcient E in terms of A,

and after some tedious but straightforward calculation we obtain

−i4kκ exp (−ik)

(κ − ik)

exp (κa) − (κ + ik)

exp (−κa)

−i2kκ exp (−ik)

(κ

− k

) sinh κa − 2iκk cosh κa

(51.7)

which after substitution into (51.6), we get

T (E) =

4κ

(κ

+ k

)

sinh

κa + 4κ

16κ

exp (−2κa)

(κ

+ k

)

(1 + exp (−4κa)) − exp (−2κa) (2κ

+ 2k

− 12κ

)

(51.8)

The above relation can be simpliﬁed for a "thick barrier", i.e. for which the

the width of the potential barrier is larger than the characteristic length scale

1/κ, so that we can neglect exp (−2κa) and exp (−4κa) in the denominator of

the expression (51.8), and it becomes

T (E) '

16κ

exp (−2κa)

(κ

+ k

)

(51.9)

or, equivalently,

(thick)

(E) ' 16





1 −



exp

−2a

2m (V

− E)

(51.10)

On the other hand, if the barrier is very thin, i.e. κa  1, the expression of

the transmission coeﬃcient in (51.8) can be approximated by

118

(thin)

(E) ' exp

−2a

2m (V

− E)

(51.11)

117

The transition current probability is deﬁned as

j(r) =

−i~

[ψ

∗

(r)∇ψ(r) −ψ(r)∇ψ

∗

(r)]

118

We used the approximation exp −2 (κa) ' 1 and exp −4 (κa) ' 1 in the denominator of the

expression (51.8).

– 79 –

We note that the transmission probability falls exponentially with the thickness

and the height of the barrier

119

• Arbitrary Potential Barrier

The above analysis can be extended to a general potential barrier by dividing it

into many small rectangular segments (see ﬁg.), so that the total transmission

probability, which we will denote by P

tunnelling

, can be obtained by taking the

product of the transmission coeﬃcients of the segments and in the thin barrier

approximation (51.11), i.e.

tunnelling

= T

.....T

' exp

−

i=1

2p(x

) δx

(51.12)

where x

is the midpoint of the ith rectangular segment of with δx

, and p(x

) =

2m (V (x

) − E)/~ is the particle’s momentum at this point. The entering

and exit of the classically forbidden region, denoted by x

and x

, respectively,

are called the turning point, since at these locations the particle’s energy is

equal to the potential energy, and hence its kinetic energy vanishes. By taking

N → ∞, or equivalently δx

→ 0, the summation in the exponential becomes

an integration and we get

tunnelling

' exp



−2

p(x) dx



= exp



−

exit

2m (V (x) − E) dx



(51.13)

• Coulomb Potential Barrier

The electrostatic potential energy of a system of two nuclei with charges Z

|e|

and Z

|e| is given by

120

V (r) =

α~cZ

(51.14)

where α ' 1/137 is the ﬁne structure constant. The two turning points are

= d

strong

∼ fm, which is the distance scale at which the strong nuclear

force operates

121

, and the point at which the particle’s energy is equal to the

electrostatic potential energy, i.e.

α~cZ

(51.15)

119

The tunneling probability of electrons is larger inside semiconductors since the electron has an

eﬀective mass smaller than its rest mass (e.g. in a semiconductor like GaAs, the eﬀective mass of

the electron is approximately 6 % of its rest mass), making the exponential less suppressed.

120

We used the relation 4π

α~c = e

121

For the α decay process, r

is the nucleus radius, which can be estimated using the semi-

empirical formula R

= 1.25A

1/3

fm based on the the liquid drop model of nucleus.

– 80 –

Then, with the use of Eq (51.13), the tunneling probability reads

tunnelling

' exp

−

strong



αZ

− E



(51.16)

which can be re-written as

tunnelling

' exp −(2γ); γ =

√

2µE

strong

− 1 dr (51.17)

Here µ = m

/(m

+ m

) is the reduced mass of the system of two nuclei. To

evaluate the above integral, we make the change of variable r = r

sin

θ, with

the integration lower and the upper limits of the integral θ

min

strong

and θ

max

= π/2, respectively, and obtain

γ =

√

2µE





− sin

−1



strong



−





strong

−



strong











(51.18)

Since r

is usually much small  d

strong

, we can make the approximation

sin

−1



strong



strong

and neglect the term (d

strong

)

, which

simpliﬁes the above expression as

γ '

√

2µE

− 2

strong

(51.19)

or, equivalently

122

γ =

παZ

√

2µc

√

+ constant (51.20)

Thus, for energies well below the Coulomb potential energy

123

, the probability

of tunneling through the barrier has the form

124

tunnelling

∝ exp

−

(51.21)

122

The constant in (51.20) is given by

constant = −2

2αµcZ

strong

= −1.17

strong

123

This is the case for energies in the core of stars.

124

The derivation of this expression is given in Appendix XX. In some nuclear physics and nuclear

astrophysics literature the tunneling probability is expressed as

tunnelling

∝ exp (−2πη(E))

where η(E) = Z

/~υ, known as the Sommerfeld parameter.

– 81 –

with υ =

2E/µ denotes the magnitude of the relative velocity of the two

nuclei, and

= (παZ

)

2µc









493 keV (51.22)

is the so-called Gamow energy.

51.2 The Rate of Nuclear Fusion in the Core of Stars

Let n

and n

be the number densities of nuclei a and b, respectively, in a plasma,

and σ the cross section of their nuclear fusion. The volumetric reaction rate for the

fusion process, i.e. the number of fusion reactions occurring per unit time per unit

volume, is given by



1 + δ



συn

(51.23)

The cross section is conventionally written as

σ(E) = S(E)σ

(geo)

tunnelling

(51.24)

where σ

(geo)

is the geometric cross section given by the area of a disk of radius equal

to the de Broglie wavelength, λ

= h/p = h/µυ, i.e.

(geo)

= πλ

= π

∝

(51.25)

The factor in S(E) in Eq (51.24), known as the astrophysical S-factor, contains

all the eﬀects due to the ﬁnite size and structure, which makes it very diﬃcult to

calculate it. Moreover, experimentally it is hard to measure the S-factor in laboratory

at low energies that correspond to temperatures inside typical stars. Instead, it is

determined by extrapolating laboratory-measured values at high energy down to

low energies. For non-resonant reactions this factor is weakly varying function of

energy

125

Since the particles are in thermal equilibrium, the expression of (51.26)

has to be folded with the Maxwell-Boltzmann distribution,

f(υ)dυ =



2πk



3/2

exp



−µυ

/2k



4πυ

dυ

√

T )

3/2

exp [−E/k

T ] dE

125

An estimate of the S-factor for proton-proton reaction is (see Ref. )

(pp)

[E] = 4 × 10

−25

MeV.barn

where barn = 10

−24

cm denotes the unit of cross section.

– 82 –

and we ﬁnd the thermally averaged rate

< R



8π



dE exp

−

exp



−



(51.26)

Note that the ﬁrst exponential increases for higher thermal velocities, whereas the

second exponential (coming from the Boltzmann distribution of velocities) decreases.

Thus, the fusion reaction rate is important only in some energy range, and it reaches

a maximum at a particular energy, E

, known as Gamow peak. The value of E

can

be determined by taking the derivative of the product of the exponential terms with

respect to energy and set it to zero, i.e.

exp

−

exp



−



= 0 (51.27)

We obtain



√



2/3

= 1.22





1/3

A[keV] (51.28)

where A = A

/(A

+ A

), with A

the atomic masses of the nuclei 1 and 2,

respectively, which is the reduced atomic mass of the system of two nuclei, and

is the temperature in million of degrees. Note that E

is larger than the average

particle’s thermal energy but much smaller than the Gamow energy of the Coulomb

barrier. Hence, for particles with energies close to E

, we expect thermonuclear

reactions to be the most eﬀective. Moreover, we can approximate the exponential

term in (51.26) with a Gaussian distribution centered around E

, i.e.

exp

−





' C exp

−



E − E

∆/2



(51.29)

Setting E = E

and using the relation between E

and E

in (51.28), we get

C = exp



−



(51.30)

For The width of the Gaussian distribution, ∆, it can be determined by taking the

second derivative of the both sides of Eq (51.29), evaluated at E = E

, which yields

∆ =

√

T (51.31)

– 83 –

51.3 Sunyaev-Zel’dovich (SZ) Eﬀect

• It’s the inverse Compton eﬀect for CMB photons against electrons in the hot

gas of clusters.

• Optical depth:

τ = nσl (51.32)

For clusters, l = few Mpc ∼ 10

cm, n < 10

−3

cm, and σ = 6.65 × 10

−25

which yields

τ ' 0.01 (51.33)

Thus, there is 1% likelihood that a CMB photon crossing the cluster is scattered

by an electron.

• Since E

 E

CMB

, an electron in the cluster transfer part of its energy to

a CMB photon that is crossing it. To ﬁrst order, this gain in energy, or,

equivalently, in frequency, is given by

∆ν

5 keV

500 keV

= 10

−2

(51.34)

• The resulting CMB anisotropy anisotropy is

∆T

=' τ

∆ν

' 10

−4

(51.35)

51.4 Speciﬁc Intensity and Radiation Pressure

51.5 Radiation Transport

51.6 Dark Matter in Cluster of Galaxies from Hot X-ray Observations

– 84 –

Figure 16: Sunyaev-Zel’dovich eﬀect on the CMB spectrum.

52 References

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